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JEE Mains 2015 10th April (online)
Physics
Single Correct Answer Type
1 In an ideal at temperature T the average force that a molecule applies on the walls of a closed
container depends on 119879 119886119904 119879119902 A good estimate for q is
(A) 2 (B) 1
2 (C) 1 (D)
1
4
Answer (C)
Solution
Average linear for collision to occur
119905 =2119889
119906
Change in momentum in 1 collision
Δ119901 = 2 119898119906
there4 average force in collision
= Δ119901
119905
119906 = root mean square speed
=2 119898119906
2119889 times 119906
rArr 119891 prop 1199062
there4 1199062 prop 119879
rArr 119891 times 119879
rArr 119902 = 1
2 In an unbiased n ndash p junction electrons diffuse from n-region to p-region because
(A) Electrons travel across the junction due to potential difference
(B) Only electrons move from n to p region and not the vice ndash versa
(C) Electron concentration in n ndash region is more as compared to that in p ndash region
(D) Holes in p ndash region attract them
Answer (C)
Solution
In a 119901 minus 119899 junction diffusion occurs due to spontaneous movement of majority charge carrier from
the region of high concentration to low concentration so option 3 in correct
3 A 10V battery with internal resistance 1Ω 119886119899119889 119886 15119881 battery with internal resistance 06Ω are
connected in parallel to a voltmeter (see figure) The reading in the voltmeter will be close to
(A) 119 119881 (B) 131 119881 (C) 125 119881 (D) 245 119881
Answer (B)
Solution
The equivalent ems of the battery combination in given as
Equation =
11986411199031 + 11986411199032
1
1199031 +
1
1199032
= 10
1 + 15
061
1 +
1
06
= 10+
150
6
1+ 10
6
=105
8
= 131 119907119900119897119905
there4 The reading measured by voltmeter = 131 119907119900119897119905
4 A proton (mass m) accelerate by a potential difference V flies through a uniform transverse
magnetic field B The field occupies a region of space by width prime119889prime 119868119891 prime120572prime be the angle of
deviation of proton from initial direction of motion (see figure) the value of sin120572 will be
(A) 119861
2radic119902119889
119898119881 (B) 119861119889radic
119902
2119898119881 (C)
119861
119889radic
119902
2119898119881 (D) 119902119881 radic
119861119889
2119898
Answer (B)
Solution
Due to potential difference V speed acquired by proton in 1199070
rArr 119882 = 119902 Δ 119881 = Δ119896
rArr 119902119907 =1
2 119898 1199070
2
rArr 1199070 = radic2119902119907
119898
Radius of circular path acquired is 119877 =1198981199070
119902119861
rArr 119877 =119898
119902119861 radic2119902119907
119898= radic
2119907119898
119902 times
1
119861
In ∆119862119875119863 sin 120572 =119889
119877= 119889radic
119902
2 119907119898 119861 = 119861119889radic
119902
2 119898119907
5 de ndash Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
(|119890| = 16 times 10minus19 119862119898119890 = 91 times 10minus31 119896119892 ℎ = 66 times 10minus34 119869119904)
(A) 05 Å (B) 12 Å (C) 17 Å (D) 24 Å
Answer (B)
Solution
De broglie wavelength 120582 in given by
120582 =ℎ
119901=
ℎ
radic2 119898119896
there4 119896119894119899119890119905119894119888 119890119899119890119903119892119910 = 119896 = 119902 Δ119907
rArr 120582 =ℎ
radic2119898119902∆119907
=66 times10minus34
radic2 times91 times 10minus3 times 16 times10minus19 times 50
=66 times10minus34
radic32 times91 times 10minus31minus19 + 2
=66 times10minus34
radic32 times91 times 10minus48
=66 times10minus34
radic5396 times 10minus24
= 122 times 10minus10
= 12 119860deg
6 Suppose the drift velocity 119907119889 in a material varied with the applied electric field E as 119907119889 prop radic119864
Then 119881 minus 119868 graph for a wire made of such a material is best given by
(A)
(B)
(C)
(D)
Answer (C)
Solution
there4 119907119889 = 119896radic119864 and 119868 = 119899 119890 119860 119907119889
rArr 119868 = 119899 119890119860 119896radic119864
there4 119864 =119907
119889 rArr 119868 = 119899119890119860119896 radic
119907
119889
rArr 119868 prop radic119907 rArr 119907 prop 1198682
So
7 A parallel beam of electrons travelling in x ndash direction falls on a slit of width d (see figure) If
after passing the slit an electron acquires momentum 119875119910 in the y ndash direction then for a majority
of electrons passing through the slit (h is Planckrsquos constant)
(A) |119875119910|119889 lt ℎ (B) |119875119910|119889 gt ℎ (C) |119875119910|119889 ≃ ℎ (D) |119875119910|119889 gt gt ℎ
Answer (D)
Solution
The electron beam will be diffractive at an angle θ
For central maxima
119889 sin 120579 = 120582
119889 sin 120579 = 119903
119901
Also 119901 sin120579 = 119901119910
rArr 119889 119901119910 = ℎ
there4 For majority of 119890120579prime119904 passing through the shit lyeing in the central maxima 119889 119901119910 asymp ℎ
8 A block of mass 119898 = 10 119896119892 rests on a horizontal table The coefficient of friction between the
block and the table is 005 When hit by a bullet of mass 50 g moving with speed v that gets
embedded in it the block moves and comes to stop after moving a distance of 2 m on the table
If a freely falling object were to acquire speed 119907
10 after being dropped from height H then
neglecting energy losses and taking 119892 = 10 119898119904minus2 the value of H is close to
(A) 02 km (B) 05 km (C) 03 km (D) 04 km
Answer ()
Solution
9 When current in a coil changes from 5 A to 2 A in 01 s an average voltage of 50 V is
produced The self ndash inductance of the coil is
(A) 167 H (B) 6 H (C) 3 H (D) 067 H
Answer (A)
Solution
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(A) 119 119881 (B) 131 119881 (C) 125 119881 (D) 245 119881
Answer (B)
Solution
The equivalent ems of the battery combination in given as
Equation =
11986411199031 + 11986411199032
1
1199031 +
1
1199032
= 10
1 + 15
061
1 +
1
06
= 10+
150
6
1+ 10
6
=105
8
= 131 119907119900119897119905
there4 The reading measured by voltmeter = 131 119907119900119897119905
4 A proton (mass m) accelerate by a potential difference V flies through a uniform transverse
magnetic field B The field occupies a region of space by width prime119889prime 119868119891 prime120572prime be the angle of
deviation of proton from initial direction of motion (see figure) the value of sin120572 will be
(A) 119861
2radic119902119889
119898119881 (B) 119861119889radic
119902
2119898119881 (C)
119861
119889radic
119902
2119898119881 (D) 119902119881 radic
119861119889
2119898
Answer (B)
Solution
Due to potential difference V speed acquired by proton in 1199070
rArr 119882 = 119902 Δ 119881 = Δ119896
rArr 119902119907 =1
2 119898 1199070
2
rArr 1199070 = radic2119902119907
119898
Radius of circular path acquired is 119877 =1198981199070
119902119861
rArr 119877 =119898
119902119861 radic2119902119907
119898= radic
2119907119898
119902 times
1
119861
In ∆119862119875119863 sin 120572 =119889
119877= 119889radic
119902
2 119907119898 119861 = 119861119889radic
119902
2 119898119907
5 de ndash Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
(|119890| = 16 times 10minus19 119862119898119890 = 91 times 10minus31 119896119892 ℎ = 66 times 10minus34 119869119904)
(A) 05 Å (B) 12 Å (C) 17 Å (D) 24 Å
Answer (B)
Solution
De broglie wavelength 120582 in given by
120582 =ℎ
119901=
ℎ
radic2 119898119896
there4 119896119894119899119890119905119894119888 119890119899119890119903119892119910 = 119896 = 119902 Δ119907
rArr 120582 =ℎ
radic2119898119902∆119907
=66 times10minus34
radic2 times91 times 10minus3 times 16 times10minus19 times 50
=66 times10minus34
radic32 times91 times 10minus31minus19 + 2
=66 times10minus34
radic32 times91 times 10minus48
=66 times10minus34
radic5396 times 10minus24
= 122 times 10minus10
= 12 119860deg
6 Suppose the drift velocity 119907119889 in a material varied with the applied electric field E as 119907119889 prop radic119864
Then 119881 minus 119868 graph for a wire made of such a material is best given by
(A)
(B)
(C)
(D)
Answer (C)
Solution
there4 119907119889 = 119896radic119864 and 119868 = 119899 119890 119860 119907119889
rArr 119868 = 119899 119890119860 119896radic119864
there4 119864 =119907
119889 rArr 119868 = 119899119890119860119896 radic
119907
119889
rArr 119868 prop radic119907 rArr 119907 prop 1198682
So
7 A parallel beam of electrons travelling in x ndash direction falls on a slit of width d (see figure) If
after passing the slit an electron acquires momentum 119875119910 in the y ndash direction then for a majority
of electrons passing through the slit (h is Planckrsquos constant)
(A) |119875119910|119889 lt ℎ (B) |119875119910|119889 gt ℎ (C) |119875119910|119889 ≃ ℎ (D) |119875119910|119889 gt gt ℎ
Answer (D)
Solution
The electron beam will be diffractive at an angle θ
For central maxima
119889 sin 120579 = 120582
119889 sin 120579 = 119903
119901
Also 119901 sin120579 = 119901119910
rArr 119889 119901119910 = ℎ
there4 For majority of 119890120579prime119904 passing through the shit lyeing in the central maxima 119889 119901119910 asymp ℎ
8 A block of mass 119898 = 10 119896119892 rests on a horizontal table The coefficient of friction between the
block and the table is 005 When hit by a bullet of mass 50 g moving with speed v that gets
embedded in it the block moves and comes to stop after moving a distance of 2 m on the table
If a freely falling object were to acquire speed 119907
10 after being dropped from height H then
neglecting energy losses and taking 119892 = 10 119898119904minus2 the value of H is close to
(A) 02 km (B) 05 km (C) 03 km (D) 04 km
Answer ()
Solution
9 When current in a coil changes from 5 A to 2 A in 01 s an average voltage of 50 V is
produced The self ndash inductance of the coil is
(A) 167 H (B) 6 H (C) 3 H (D) 067 H
Answer (A)
Solution
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Solution
Due to potential difference V speed acquired by proton in 1199070
rArr 119882 = 119902 Δ 119881 = Δ119896
rArr 119902119907 =1
2 119898 1199070
2
rArr 1199070 = radic2119902119907
119898
Radius of circular path acquired is 119877 =1198981199070
119902119861
rArr 119877 =119898
119902119861 radic2119902119907
119898= radic
2119907119898
119902 times
1
119861
In ∆119862119875119863 sin 120572 =119889
119877= 119889radic
119902
2 119907119898 119861 = 119861119889radic
119902
2 119898119907
5 de ndash Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
(|119890| = 16 times 10minus19 119862119898119890 = 91 times 10minus31 119896119892 ℎ = 66 times 10minus34 119869119904)
(A) 05 Å (B) 12 Å (C) 17 Å (D) 24 Å
Answer (B)
Solution
De broglie wavelength 120582 in given by
120582 =ℎ
119901=
ℎ
radic2 119898119896
there4 119896119894119899119890119905119894119888 119890119899119890119903119892119910 = 119896 = 119902 Δ119907
rArr 120582 =ℎ
radic2119898119902∆119907
=66 times10minus34
radic2 times91 times 10minus3 times 16 times10minus19 times 50
=66 times10minus34
radic32 times91 times 10minus31minus19 + 2
=66 times10minus34
radic32 times91 times 10minus48
=66 times10minus34
radic5396 times 10minus24
= 122 times 10minus10
= 12 119860deg
6 Suppose the drift velocity 119907119889 in a material varied with the applied electric field E as 119907119889 prop radic119864
Then 119881 minus 119868 graph for a wire made of such a material is best given by
(A)
(B)
(C)
(D)
Answer (C)
Solution
there4 119907119889 = 119896radic119864 and 119868 = 119899 119890 119860 119907119889
rArr 119868 = 119899 119890119860 119896radic119864
there4 119864 =119907
119889 rArr 119868 = 119899119890119860119896 radic
119907
119889
rArr 119868 prop radic119907 rArr 119907 prop 1198682
So
7 A parallel beam of electrons travelling in x ndash direction falls on a slit of width d (see figure) If
after passing the slit an electron acquires momentum 119875119910 in the y ndash direction then for a majority
of electrons passing through the slit (h is Planckrsquos constant)
(A) |119875119910|119889 lt ℎ (B) |119875119910|119889 gt ℎ (C) |119875119910|119889 ≃ ℎ (D) |119875119910|119889 gt gt ℎ
Answer (D)
Solution
The electron beam will be diffractive at an angle θ
For central maxima
119889 sin 120579 = 120582
119889 sin 120579 = 119903
119901
Also 119901 sin120579 = 119901119910
rArr 119889 119901119910 = ℎ
there4 For majority of 119890120579prime119904 passing through the shit lyeing in the central maxima 119889 119901119910 asymp ℎ
8 A block of mass 119898 = 10 119896119892 rests on a horizontal table The coefficient of friction between the
block and the table is 005 When hit by a bullet of mass 50 g moving with speed v that gets
embedded in it the block moves and comes to stop after moving a distance of 2 m on the table
If a freely falling object were to acquire speed 119907
10 after being dropped from height H then
neglecting energy losses and taking 119892 = 10 119898119904minus2 the value of H is close to
(A) 02 km (B) 05 km (C) 03 km (D) 04 km
Answer ()
Solution
9 When current in a coil changes from 5 A to 2 A in 01 s an average voltage of 50 V is
produced The self ndash inductance of the coil is
(A) 167 H (B) 6 H (C) 3 H (D) 067 H
Answer (A)
Solution
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(A)
(B)
(C)
(D)
Answer (C)
Solution
there4 119907119889 = 119896radic119864 and 119868 = 119899 119890 119860 119907119889
rArr 119868 = 119899 119890119860 119896radic119864
there4 119864 =119907
119889 rArr 119868 = 119899119890119860119896 radic
119907
119889
rArr 119868 prop radic119907 rArr 119907 prop 1198682
So
7 A parallel beam of electrons travelling in x ndash direction falls on a slit of width d (see figure) If
after passing the slit an electron acquires momentum 119875119910 in the y ndash direction then for a majority
of electrons passing through the slit (h is Planckrsquos constant)
(A) |119875119910|119889 lt ℎ (B) |119875119910|119889 gt ℎ (C) |119875119910|119889 ≃ ℎ (D) |119875119910|119889 gt gt ℎ
Answer (D)
Solution
The electron beam will be diffractive at an angle θ
For central maxima
119889 sin 120579 = 120582
119889 sin 120579 = 119903
119901
Also 119901 sin120579 = 119901119910
rArr 119889 119901119910 = ℎ
there4 For majority of 119890120579prime119904 passing through the shit lyeing in the central maxima 119889 119901119910 asymp ℎ
8 A block of mass 119898 = 10 119896119892 rests on a horizontal table The coefficient of friction between the
block and the table is 005 When hit by a bullet of mass 50 g moving with speed v that gets
embedded in it the block moves and comes to stop after moving a distance of 2 m on the table
If a freely falling object were to acquire speed 119907
10 after being dropped from height H then
neglecting energy losses and taking 119892 = 10 119898119904minus2 the value of H is close to
(A) 02 km (B) 05 km (C) 03 km (D) 04 km
Answer ()
Solution
9 When current in a coil changes from 5 A to 2 A in 01 s an average voltage of 50 V is
produced The self ndash inductance of the coil is
(A) 167 H (B) 6 H (C) 3 H (D) 067 H
Answer (A)
Solution
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(A) |119875119910|119889 lt ℎ (B) |119875119910|119889 gt ℎ (C) |119875119910|119889 ≃ ℎ (D) |119875119910|119889 gt gt ℎ
Answer (D)
Solution
The electron beam will be diffractive at an angle θ
For central maxima
119889 sin 120579 = 120582
119889 sin 120579 = 119903
119901
Also 119901 sin120579 = 119901119910
rArr 119889 119901119910 = ℎ
there4 For majority of 119890120579prime119904 passing through the shit lyeing in the central maxima 119889 119901119910 asymp ℎ
8 A block of mass 119898 = 10 119896119892 rests on a horizontal table The coefficient of friction between the
block and the table is 005 When hit by a bullet of mass 50 g moving with speed v that gets
embedded in it the block moves and comes to stop after moving a distance of 2 m on the table
If a freely falling object were to acquire speed 119907
10 after being dropped from height H then
neglecting energy losses and taking 119892 = 10 119898119904minus2 the value of H is close to
(A) 02 km (B) 05 km (C) 03 km (D) 04 km
Answer ()
Solution
9 When current in a coil changes from 5 A to 2 A in 01 s an average voltage of 50 V is
produced The self ndash inductance of the coil is
(A) 167 H (B) 6 H (C) 3 H (D) 067 H
Answer (A)
Solution
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Area of coil
119889 = 119871119868 rArr ∆119889
∆119905= 119871
∆119868
∆119905
there4 (120576119894119899119889)119886119907119890119903119886119892119890 = |∆119889
∆119905| = 119871 |
∆119868
∆119905|
rArr 50 = 119871 times 5minus2
01
rArr 5
3= 119871
rArr 119871 = 1674
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
trajectory will look like
(A)
(B)
(C)
(D)
Answer (C)
Solution
∵ 119909 = 119860 sin120596119905 rArr 119904119894119899 120596119905 =119909
119860
Also 119888119900119904 120596119905 = radic1 minus sin2120596119905 = radic1 minus1199092
1198602
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
rArr cos 120596119905 = radic1198602minus1199092
119860
As 119910 = 2119860 sin120596119905 cos120596119905
rArr 119910 = 2 119860119909
119860 radic1198602 minus 1199092
119860
rArr 119910 =2
119860 119909 radic1198602 minus 1199092
rArr 119910 = 0 119886119905 119909 = 0 119886119899119889 119909 = plusmn 119860
Which in possible only in option (3)
11 Consider a thin uniform square sheet made of a rigid material If its side is lsquoarsquo mass m and
moment of inertia I about one of its diagonals then
(A) 119868 =1198981198862
24
(B) 1198981198862
24lt 119868 lt
1198981198862
12
(C) 119868 gt1198981198862
12
(D) 119868 =1198981198862
12
Answer (D)
Solution
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through
centre and lying in the blank of the plate is same
there4 119868119889119894119886119892119900119899119886119897 = 119868119901119886119903119886119897119897119890119897 119905119900 119904119894119889119890
=1198981198862
12
12 Diameter of a steel ball is measured using a Vernier calipers which has divisions of 01 cm on
its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main
scale Three such measurements for a ball are given as
SNo MS (cm) VS divisions 1 05 8 2 05 4 3 05 6
If the zero error is ndash 003 cm then mean corrected diameter is
(A) 053 cm
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(B) 056 cm
(C) 059 cm
(D) 052 cm
Answer (C)
Solution
LC of Vernier calipers
= 1 119898119886119894119899 119904119888119886119897119890 119889119894119886119898119890119905119890119903
119879119900119905119886119897 119889119894119907119894119904119900119899119904 vernier 119904119888119886119897119890
=01
10= 001 119888119898
Required of Vernier calipers
= 119872 119878 119877+(119871 119862) times 119907119904 119889119894119907119894119904119894119900119899119904
there4 Measured diameter are respecting
052 119888119898 054 119888119898 056 119888119898
there4 119886119907119890119903119886119892119890 119889119894119886119898119890119905119890119903 = 058 + 054 + 056
3
=168
3= 056
there4 119888119900119903119903119890119888119905119890119889 119889119894119886119898119890119905119890119903 = 056 minus (minus003)
= 056 + 003 = 059 119888119898
13 A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
R (R lt lt L) A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre If the time period of star is T and its distance from the
galaxyrsquos axis is r then
(A) 119879 prop radic119903
(B) 119879 prop 119903
(C) 119879 prop 1199032
(D) 1198792 prop 1199033
Answer (B)
Solution
Due to a long solid cylinder gravitational field strong can be given as
119892prime = 2 119866 120582
119909
Where
120582 = 119897119894119899119890119886119903 119898119886119904119904 119889119890119899119904119894119905119910 119900119891 119892119886119897119886119909119910
119865119900119903 119905ℎ119890 119900119903119887119894119905119886119897 119898119900119905119894119900119899 119886119903119900119906119899119889 119905ℎ119890 119892119886119897119886119909119910
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119891119892 = 119891119888119890119899119905119903119894119901119890119905119886119897
rArr 119898119892 = 119898 1205962119909
rArr 2119866120582
119909= 1205962119909
rArr 1205962 prop1
1199092
rArr 120596 prop1
119909
rArr 2120587
119879 prop
1
119909 rArr 119879 prop 119909
So option 2 is correct
14 An electromagnetic wave travelling in the x ndash direction has frequency of 2 times 1014 119867119911 and
electric field amplitude of 27 119881119898minus1 From the options given below which one describes the
magnetic field for this wave
(A) (119909 119905) = (9 times 10minus8119879)119895 sin[15 times 10minus6 119909 minus 2 times 1014119905]
(B) (119909 119905) = (9 times 10minus8119879)119894 sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(C) (119909 119905) = (3 times 10minus8119879) sin[2120587(15 times 10minus8 119909 minus 2 times 1014119905)]
(D) (119909 119905) = (9 times 10minus8119879) sin[2120587 (15 times 10minus6 119909 minus 2 times 1014119905)]
Answer (D)
Solution
119882ℎ119890119899 119864 = 1198640 119904119894119899 119862 119896119909 minus 120596119905
119879ℎ119890119899 119861 = 1198610 119904119894119899 119862 119896119909 minus 120596119905
Of light in travelling along 119894 then in either along 119895 or
there4 119878119901119890119890119889 119900119891 119897119894119892ℎ119905 119862 = 1198640
1198610 rArr 1198610 =
1198640
119862
rArr 1198610 = 27
3times108= 9 times 10minus8 119879
also 120596 = 2120587 f = 2π times 2 times 1014 = 4 120587 times 1014
Looking into the option the correct
Answer is = 9 times 10minus8 sin2120587 (15 times 10minus6119909 minus 2 times 1014119905)
15 A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm If
a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting the
angle formed by the image of the tower is 120579 then 120579 is close to
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(A) 30deg
(B) 15deg
(C) 1deg
(D) 60deg
Answer (D)
Solution
16 A block of mass 119898 = 01 119896119892 is connected to a spring of unknown spring constant k It is
compressed to a distance x from its equilibrium position and released from rest After
approaching half the distance (119909
2) from equilibrium position it hits another block and comes
to rest momentarily while the other block moves with a velocity 3 119898119904minus1 The total initial
energy of the spring is
(A) 06 119869
(B) 08 119869
(C) 15 119869
(D) 03 119869
Answer (A)
Solution By energy conservation between compression positions 119909 and 119909
2
1
21198961199092 =
1
2119896 (119909
2)2
+1
21198981199072
1
21198961199092 minus
1
21198961199092
4=1
21198981199072
1
21198961199092 (
3
4) =
1
21198981199072
119907 = radic31198961199092
4119898= radic
3119896
119898
119909
2
On collision with a block at rest
∵ Velocities are exchanged rArr elastic collision between identical masses
there4 119907 = 3 = radic3119896
119898
119909
2
rArr 6 = radic3119896
119898 119909
rArr 119909 = 6radic119898
3119896
there4 The initial energy of the spring is
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119880 =1
2119896 1199092 =
1
2119896 times 36
119898
3119896= 6119898
119880 = 6 times 01 = 06 119869
17 Shown in the figure are two point charges + Q and ndash Q inside the cavity of a spherical shell The
charges are kept near the surface of the cavity on opposite sides of the centre of the shell If 1205901is
the surface charge on the inner surface and 1198761net charge on it and 1205902 the surface charge on the
other surface and 1198762 net charge on it then
(A) 1205901 = 01198761 = 0 1205902 = 01198762 = 0
(B) 1205901 ne 01198761 = 0 1205902 ne 01198762 = 0
(C) 1205901 ne 01198761 ne 0 1205902 ne 01198762 ne 0
(D) 1205901 ne 01198761 = 0 1205902 = 01198762 = 0
Answer (D)
Solution By the property of electrostatic shielding in the conductors 120598 = 0 in the conductor
So electric flux = 0 through a dotted Gaussian surface as shown
The net enclosed charge through Gaussian surface = 0
rArr Net charge 1198761 on the inner surface = 0 but the equal and opposite induced charge on the surface
will be distributed non uniformly on the inner surface
So 1205901 ne 0
∵ 1198761 = 0 on the inner surface
So net charge 1198762 = 0 on the outer surface as conductor is neutral but ∵ outer surface is free from
any electric field so no charge density exists on the outer surface So 1205902 = 0
18 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face
and views the magnified image of the face at the closest comfortable distance of 25 cm The
radius of curvature of the mirror would then be
(A) 24 119888119898
(B) 30 119888119898
(C) 60 119888119898
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(D) minus24 119888119898
Answer (C)
Solution
If AB is the position of face of man then A lsquoBrsquo is the position of image of face
As image is formed at 25cm form the object
there4 From concave mirror image is 15cm behind the mirror
So 119906 = minus10 119888119898 119907 = +15 119888119898
rArr1
119891=1
119906+1
119907
rArr1
119891=
1
minus10+1
15=minus3 + 2
30
rArr 119891 = minus300 119888119898
So radius of curvature = 60 119888119898
19 A thin disc of radius 119887 = 2119886 has a concentric hole of radius lsquoarsquo in it (see figure) It carries
uniform surface charge prime120590prime on it If the electric field on its axis at height primeℎprime(ℎ lt lt 119886) from its
centre is given as lsquoChrsquo then value of lsquoCrsquo is
(A) 120590
4 1205721205980
(B) 120590
1205721205980
(C) 120590
1198781205721205980
(D) 120590
21205721205980
Answer (A)
Solution ∵ at the axial point of a uniformly charged disc electric field is given by
119864 =120590
21205980(1 minus 119888119900119904120579)
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
By superposition principle when inner disc is removed then electric field due to remaining disc is
119864 =120590
21205980 [(1 minus 1198881199001199041205792) minus (1 minus 1198881199001199041205791)]
=120590
21205980[1198881199001199041205791 minus 1198881199001199041205792]
=120590
21205980[
ℎ
radicℎ2 + 1198862 minus
ℎ
radicℎ2 + 1198872]
=120590
21205980[
ℎ
119886radic1 +ℎ2
1198862
minusℎ
radic1 +ℎ2
1198872 ]
∵ ℎ ≪ 119886 and b
there4 119864 =120590
21205980[ℎ
119886minusℎ
119887]
=120590
21205980[ℎ
119886minusℎ
2119886] =
120590ℎ
41205980119886
rArr 119862 =120590
41198861205980
20 An ideal gas goes through a reversible cycle 119886 rarr 119887 rarr 119888 rarr 119889 has the V ndash T diagram shown below
Process 119889 rarr 119886 119886119899119889 119887 rarr 119888 are adiabatic
The corresponding P ndash V diagram for the process is (all figures are schematic and not drawn to
scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(B)
(C)
(D)
Answer (A)
Solution Is an adiabatic process
119879119881120574minus1 = 119888119900119899119904119905 rArr 1198811198791120574minus1 = 119888119900119899119904119905
rArr as T increase V decreases at non-uniform rate
In process 119886 rarr 119887 P = constant as 119881 prop 119879
In process 119888 rarr 119889 119875prime = constant s 119881 prop 119879
But since slope of V ndash T graph prop1
119875
since slope of ab lt slope of cd
rArr 119875119886119887 gt 119875119888119889
Also in adiabatic process 119889 rarr 119886 as T is increasing V in decreasing
rArr P is increasing so P ndash V diagram is as below
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
21 A uniform solid cylindrical roller of mass lsquomrsquo is being pulled on a horizontal surface with force F
parallel to the surface and applied at its centre If the acceleration of the cylinder is lsquoarsquo and it is
rolling without slipping then the value of lsquoFrsquo is
(A) 3
2 119898119886
(B) 2 119898119886
(C) 5
3 119898119886
(D) 119898119886
Answer (A)
Solution
From free body diagram of cylinder
119865 minus 119891119904 = 119898119886 hellip(1)
∵ sum 119891119890119909119905 = 119898119886119888119898
119886119897119904119900 sum 120591119890119909119905 = 119868119888119898 prop
⟹ 119891119904 119877 = 119868119888119898 prop
⟹ 119891119904 119877 =1
2 1198981198772 prop hellip (2)
For rolling without slipping
119886 = 119877 prop helliphellip (3)
⟹ prop=119902
119877
there4 119891119904 119877 =1
21198981198772
119902
119877
⟹ 119891119904 =1
2119898119886
Put in (1)
119891 minus1
2119898119886 = 119898119886
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
⟹ 119891 =3
2119898119886
22 A 25 cm long solenoid has radius 2 cm and 500 total number of turns It carries a current of 15
A If it is equivalent to a magnet of the same size and magnetization
(119872119886119892119899119890119905119894119888 119898119900119898119890119899119905 119881119900119897119906119898119890frasl ) 119905ℎ119890119899 | | is
(A) 3120587 119860119898minus1
(B) 30000 119860119898minus1
(C) 30000120587 119860119898minus1
(D) 300 119860119898minus1
Answer (B)
Solution
119881119900119897119906119898119890 = 119860119897
119872119886119892119899119890119905119894119911119886119905119894119900119899 =119898119886119892119899119890119905119894119888 119898119900119899119890119899119905
119881119900119897119906119898119890
=(119873119900119900119891 119905119906119903119899119904)times(119862119906119903119903119890119899119905)times119860119903119890119886
119881119900119897119906119898119890
=119873 119868 119860
119860 ℓ
=119873119868
ℓ
=500times15times100
25
= 60 times 500
= 30 times 103
= 30000 119860119898minus1
23 In the circuits (a) and (b) switches 1198781 119886119899119889 1198782 are closed at t = 0 and are kept closed for a long
time The variation of currents in the two circuits for 119905 ge 0 are roughly shown by (figures are
schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(B)
(C)
(D)
Answer (B)
Solution
In CR series circuit
119902 = 1199020 (1 minus 119890minus119905
120591 )
⟹ 119902 = 119862119864 (1 minus 119890minus119905
119877119862)
there4 119862119906119903119903119890119899119905 119868 =119889119902
119889119905
=119862119864
119877119862(+119890
minus119905
119877119862)
119868 =119864
119877 119890minus119905
119877119862
⟹ 119888119906119903119903119890119899119905 119889119890119888119886119910119904 119890119909119901119900119899119890119899119905119894119886119897119897119910 119886119889 119894119899 119871119877 119904119890119903119894119890119904 119888119894119903119888119906119894119905
119868 = 1198680 (1 minus 119890minus119905
120591 )
119908ℎ119890119903119890 1198680 =119864
119877 119886119899119889 120591 =
119871
119877
119868 =119864
119877 (1 minus 119890
minus119877119905
119871 ) ⟹ 119888119906119903119903119890119899119905 119892119903119900119908119904 119890119909119901119900119899119890119899119905119894119886119897119897119904
there4 119891119900119903 119862 minus 119877 119888119894119903119888119906119894119905
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
For L ndash R circuit
24 If two glass plates have water between them and are separated by very small distance (see
figure) it is very difficult to pull them apart It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison to
atmosphere If the radius of the cylindrical surface is R and surface tension of water is T then
the pressure in water between the plates is lower by
(A) 2119879
119877
(B) 119879
4119877
(C) 4119879
119877
(D) 119879
2119877
Answer (A)
Solution
119889 = 2119877 119888119900119904120579
there4 119875119903119890119904119904119906119903119890 119889119894119891119891119890119903119890119899119888119890 119886119888119903119900119904119904 119886 119889119900119906119887119897119890 119888119906119903119907119886119905119906119903119890 119891119894119897119898
∆119875 = 2119879 (1
1198771+
1
1198772)
∵ 1198771 = 119877 119886119899119889 1198772 = infin
∆119875 = 2119879 (1
119877+1
infin)
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
∆119875 = 2119868
119877
there4 Pressure is more in the concave side hence pressure in water between the plates is lower by 2119879
119877
25 A simple harmonic oscillator of angular frequency 2 rad 119904minus1 is acted upon by an external force
119865 = sin 119905 119873 If the oscillator is at rest in its equilibrium position at 119905 = 119900 its position at later
times is proportional to
(A) sin 119905 +1
2cos 2119905
(B) 119888119900119904119905 minus1
2sin2119905
(C) sin 119905 minus1
2sin2119905
(D) sin 119905 +1
2sin2119905
Answer (C)
Solution
It is given that oscillator at rest at t = 0 ie at t = 0 v = 0
So in option we can check by putting 119907 =119889119909
119889119905= 0
(1) 119868119891 119909 prop sin 119905 +1
2cos2119905
⟹ 119907 prop cos 119905 +1
2times 2 (minus sin 2119905)
⟹ 119886119905 119905 = 0 119907 prop 1 minus 0 ne 0
(2) 119868119891 119909 prop cos 119905 minus1
2sin 119905
⟹ 119907 prop minus sin 119905 minus1
2cos 119905
⟹ 119886119905 119905 = 0 119907 prop minus1
2ne 0
(3) 119868119891 119909 prop sin 119905 minus1
2 119904119894119899120579 2119905
119905ℎ119890119899 120592 prop cos 119905 minus1
2times 2 cos 2119905
⟹ 119886119905 119905 = 0 119907 prop 1 minus 1 = 0
(4) 119868119891 119909 prop sin 119905 +1
2sin2119905
⟹ 119907 prop cos 119905 +1
2times 2 cos2119905
⟹ 119886119905 119905 = 0 119907 prop 1 + 1
⟹ 119907 prop 2 ne 0
there4 119894119899 119900119901119905119894119900119899 (3) 119907 = 0 119886119905 119905 = 0
26 If a body moving in a circular path maintains constant speed of 10 119898119904minus1 then which of the
following correctly describes relation between acceleration and radius
(A)
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(B)
(C)
(D)
Answer (D)
Solution
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
119886 =1199072
119877
⟹ 119886119877 = 119888119900119899119904119905119886119899119905
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
For a charge q moving in a +r uniform magnetic field B
119891119898 =1198981199072
119877
119902119881119861 = 1198981199072
119877
rArr 1198981199072 = 119902119881119861119877
rArr 1
2 1198981199072 =
119902119881119861119877
2
rArr 119864119899119890119903119892119910 =119902119881119861119877
2 (1)
By Bohrrsquos quantisation condition
Angular momentum 119871 = 119899ℎ
2120587
rArr 119898119907119877 =119899ℎ
2120587
rArr 119907119877 =119899ℎ
2120587 119898 (2)
Put (2) in (2)
rArr 119864119899119890119903119892119910 =119902119861
2 (
ℎ
2 120587 119898)
= 119902119861 119899ℎ
4 120587 119898
29 If the capacitance of a nanocapacitor is measured in terms of a unit lsquoursquo made by combining the
electronic charge lsquoersquo Bohr radius prime1198860prime Planckrsquos constant lsquohrsquo and speed of light lsquocrsquo then
(A) 119906 =11989021198860
ℎ119888 (B) 119906 =
ℎ119888
11989021198860 (C) 119906 =
1198902119888
ℎ1198860 (D) 119906 =
1198902ℎ
1198881198860
Answer (A)
Solution
∵ 119862119886119901119886119888119894119905119886119899119888119890 119862 =119876
∆119907
119860119897119904119900 [ℎ119888
120582] = [
ℎ119888
1198860] = [119864119899119890119903119892119910]
there4 [119862] = [119876]
[∆119907]=
[119876] [119876]
[∆119907] [119876]
∵ 119882 = 119902∆119907 rArr [119876] [∆119907] = [119864119899119890119903119892119910]
there4 [119862] = [1198762]
[119864119899119890119903119892119910]=
[1198762] [1198860]
[ℎ119888]
there4 [119862119886119901119886119888119894119905119886119899119888119890 ] = [1198762] [1198860]
[ℎ119888]
rArr 119906 = 1198902 1198860
ℎ119888
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
30 A bat moving at 10 119898119904minus1 towards a wall sends a sound signal of 8000 Hz towards it On
reflection it hears a sound of frequency119891 The value of 119891 in Hz is close to
(119904119901119890119890119889 119900119891 119904119900119906119899119889 = 320 119898119904minus1)
(A) 8258
(B) 8424
(C) 8000
(D) 8516
Answer (D)
Solution
We can assume that reflected wave is due to image of B coming with same speed in opposite
direction
Observer
119891 = 119907 + 10
119907 minus 10 times 1198910
= 320 + 10
320 minus 10 times 8000
= 330
310 times 8000
= 33
31 times 8000
= 8516 119867119911
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
JEE Mains 2015 10th April (online)
Chemistry
Single correct answer type
1 14 g of an organic compound was digested according to Kjeldahlrsquos method and the ammonia
evolved was absorbed in 60 mL of M10 11986721198781198744 solution The excess sulphuric acid required 20
mL of M10 NaOH solution for neutralization The percentage of nitrogen in the compound is
(A) 24 (B)3 (C)5 (D)10
Solution (D) 60 times1
10= 6 119898119872 11986721198781198744 used
Excess 11986721198781198744 equiv 20 times1
10times1
2= 1 119898119872 11986721198781198744
11986721198781198744 used = 6 minus 1 = 5 119898119872
21198731198673 +11986721198781198744⟶ (1198731198674)21198781198744
mM of 1198731198673 = 10 119898119872
Mass of 119873 = 10 times 10minus3 times 14 (119892
119898119900119897119890) = 0140119892
1198732 =0140
14times 100 = 10
2 The optically inactive compound from the following is
(A) 2-chloropropanal
(B) 2-chloro-2-methylbutane
(C) 2-chlorobutane
(D) 2-chloropentane
Solution (B)
(Optically active)
(Optically inactive because of 2 minus 1198621198673 groups present on same C atom)
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(Optically active)
3 The least number of oxyacids are formed by
(A) Chlorine
(B) Fluorine
(C) Sulphur
(D) Nitrogen
Solution (B) Fluorine does not form oxyacids as it is more electronegative than oxygen
4 Gaseous 11987321198744 dissociates into gaseous 1198731198742according to the reaction11987321198744(119892) 21198731198742(119892)
At 300 K and 1 atm pressure the degree of dissociation of 11987321198744 is 02 If one mole of 11987321198744 gas is
contained in a vessel then the density of the equilibrium mixture is
(A) 311 gL
(B) 156 gL
(C) 456 gL
(D) 622 gL
Solution (A) 11987321198744 21198731198742(1 minus 120572) 2120572
Total moles at equilibrium = 1 minus 120572 + 2120572 = 1 + 120572 = 12
M avg for equilibrium mixture =92
119892
119898119900119897119890 (11987321198744)
12
119889119886119907119890119903119886119892119890 =119875119872119886119907119892
119877119879=
1 times 7667
0082 times 300=7667
246
= 311 119892119871minus1
5 Arrange the following amines in the order of increasing basicity
(A)
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(B)
(C)
(D)
Solution (C)
Most basic due to +I effect of methyl group Methoxy group provides electron density at -
1198731198672
-1198731198742 group with draws electron density from N of -1198731198672
6
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
A is
(A)
(B)
(C)
(D)
Solution (A)
7 A solution at 20119900119862 is composed of 15 mol of benzene and 35 mol of toluene If the vapour
pressure of pure benzene and pure toluene at this temperature are 747 torr and 223 torr
respectively then the total vapour pressure of the solution and the benzene mole fraction in
equilibrium with it will be respectively
(A) 305 torr and 0389
(B) 350 torr and 0480
(C) 380 torr and 0589
(D) 358 torr and 0280
Solution (C) 119883119861119890119899119911119890119899119890 =15
5= 03
119883119879119900119897119906119890119899119890 =35
5= 07
119875119905119900119905119886119897 = 03 times 747 + 07 times 223
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
= 2241 + 1561 = 3802
asymp 38 119879119900119903119903
By Daltonrsquos law to vapour phase
119883119861119890119899119911119890119899119890prime (119907119886119901 119901ℎ119886119904119890) =
03 times 747
38=2241
38
= 0589
8 Which moleculeion among the following cannot act as a ligand in complex compounds
(A) 119862119873minus
(B) 1198621198674
(C) 119862119874
(D) 119861119903minus
Solution (B) 1198621198674 does not have either a lone pair or 120587-electron pair it cannot act as ligand
9 A compound A with molecular formula 1198621011986713119862119897 gives a white precipitate on adding silver
nitrate solution A on reacting with alcoholic KOH gives compound B as the main product B on
ozonolysis gives C and D C gives Cannizaro reaction but not aldol condensation D gives aldol
condensation but not Cannizaro reaction A is
(A)
(B)
(C)
(D)
Solution (B) Chlorine attached to tertiary carbon will give a white precipitate on adding 1198601198921198731198743
(Saytzeff Rule)
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
10
is used as
(A) Antacid
(B) Insecticide
(C) Antihistamine
(D) Analgesic
Solution (D) Acetyl salicylic acid is analgesic
11 An aqueous solution of a salt X turns blood red on treatment with 119878119862119873minus and blue on
treatment with 1198704[119865119890(119862119873)6] X also gives a positive chromyl chloride test The salt X is
(A) 1198651198901198621198973
(B) 119865119890(1198731198743)3
(C) 1198621199061198621198972
(D) 119862119906(1198731198743)2
Solution (A) 1198651198901198621198713 + 3 119878119862119873119886119902
minus 119865119890(119878119862119873)3 + 3 119862119897minus
(119861119897119900119900119889 119903119890119889)
4 1198651198901198621198973 + 31198704[119865119890(119862119873)6] ⟶ 12 119870119862119897 + 1198651198904[119865119890(119862119873)6]3119875119903119906119904119904119894119900119899 119887119897119906119890
21198651198901198621198973 + 311986721198781198744⟶ 1198651198902(1198781198744)3 + 6119867119862119897
119870211986211990321198747 + 211986721198781198744⟶ 21198701198671198781198744 + 21198621199031198743 +1198672119874 1198621199031198743 + 2119867119862119897 ⟶ 11986211990311987421198621198972 +1198672119874
(119862ℎ119903119900119898119910119897119888ℎ119897119900119903119894119889119890)
11986211989011987421198621198972 + 4 119873 119886119874119867 ⟶ 11987311988621198621199031198744 + 2119873119886119862119897 + 21198672119874
(119910119890119897119897119900119908)
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
11987311988621198621199031198744 + 119875119887(1198621198673119862119874119874)2 ⟶ 1198751198871198621199031198744 + 21198621198673119862119874119874119873119886
(119910119890119897119897119900119908 119901119901119905)
12 The correct statement on the isomerism associated with the following complex ions
(A) [119873119894(1198672119874)51198731198673]2+
(B) [119873119894(1198672119874)4(1198731198673)2]2+ and
(C) [119873119894(1198672119874)3(1198731198673)3]2+ is
(D) (A) and (B) show only geometrical isomerism
Solution (D) [119873119894 (1198672119874)4(1198731198673)2]2+
Show c is amp trans geometrical isomerism [119873119894 (1198672119874)3(1198731198673)3]2+
Show facial amp meridional geometrical isomerism
13 In the presence of a small amount of phosphorous aliphatic carboxylic acids react with 120572-
hydrogen has been replaced by halogen This reaction is known as
(A) Etard reaction
(B) Wolff-Kischner reaction
(C) Rosenmund reaction
(D) Hell-volhard-zelinsky reaction
Solution (D) This reaction is known as HVZ reaction
14 The reaction 2N2O5(g) rarr 4NO2(g) + O2(g) follows first order kinetics The pressure of a
vessel containing only N2O5 was found to increase from 50 mm Hg to 875 mm Hg in 30 min
The pressure exerted by the gases after 60 min Will be (Assume temperature remains
constant)
(A) 10625 mm Hg
(B) 125 mm Hg
(C) 11625 mm Hg
(D) 150 mm Hg
Solution (A) 211987321198745(119892) ⟶ 4 1198731198742(119892)
(1199010 minus 119909) 2119909 + 1198742(119892)
119909
2
sum119901119903119890119904119904119906119903119890 =1199010 minus 119909 + 2119909 +119909
2= 1199010 +
3119909
2= 119901119905119900119905119886119897
875 = 50 +3119909
2
3119909
2= 375
there4 119909 = 375 times2
3= 25
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
For first order kinetics
119896119905 = ln1199010
1199010 minus 119909= 119897119899
50
25= ln2
119896 =1
119905ln 2 =
1
30ln 2
After 60 min
119896 =1
119905primeln
11990101199010 minus 119909
primerArr1
30ln 2 =
1
60ln
11990101199010 minus 119909
prime
2 ln 2 = ln1199010
1199010 minus 119909primeminus ln 4
11990101199010 minus 119909
prime= 4 rArr 1199010 = 4 1199010 minus 4119909prime
119909prime =41199010 minus 1199010
4=311990104=3 times 50
4= 375
Σ60 119898119894119899119879119900119905119886119897 119901119903119890119904119904119906119903119890 = 1199010 +3119909prime
2= 50 + 3 times
375
2
= 50 + 5625 = 10625 119898119898
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
1 minus1
2 minus16radic2
3 minus8
4 minus2radic2
Answer (3)
Solution |11990911 11199101 11119911| = 119909119910119911 minus (119909 + 119910 + 119911) + 2
Since 119860119872 ge 119866119872
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119909 + 119910 + 119911
3 ge (119909119910119911)
13
119909 + 119910 + 119911 ge 3(119909119910119911)13
there4 Least value of xyz will have from (when determinant non- negative terms)
119909119910119911 minus (3)(119909119910119911)13 + 2 ge 0
1199053 minus 3119905 + 2 ge 0
(119905 + 2)(1199052 minus 2119905 + 1)
119905 = minus2 119886119899119889 119905 = +1
Least value of 1199053 = minus8
3 The contrapositive of the statement ldquoIf it is raining then I will not comerdquo is
1 If I will come then it is not raining 2 If I will come then it is raining
3 If I will not come then it is raining
4 If I will not come then it is not raining
Answer (1)
Solution Contrapositive of 119875 rArr 119902 is
~119902 rArr ~ 119875 So contra positive of the statement ldquoIf it is raining then I will not comerdquo would be
If I will come then it is not raining
4 lim119909rarr0
1198901199092minuscos119909
sin2 119909 is equal to
1 2
2 3
2
3 5
4
4 3
Answer (2)
Solution 1198901199092minuscos119909
sin2 119909=
(1 + 1199092
∟1 + 1199094
∟2helliphellip) minus (1 minus 1199092
∟2 + 1199094
∟4helliphellip119899)
sin2 1199091199092
minus 1199092
(+31199092
2+11 1199094
24sin2 119909
1199092 ∙1199092) take 1199092 common
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
[lim119909rarr0
+32 +
1124 119909
2
sin2 1199091199092
] =3
2
5 If Rollersquos theorem holds for the function 119891(119909) = 21199093 + 1198871199092 + 119888119909 119909 isin [minus1 1] at the point 119909 =1
2
then 2b + c equals
1 2
2 1
3 -1
4 -3
Answer (3)
Solution If Rollersquos theorem is satisfied in the interval [-1 1] then
119891(minus1) = 119891(1)
minus2 + 119887 minus 119888 = 2 + 119887 + 119888
119888 = minus2 also 119891prime(119909) = 61199092 + 2119887119909 + 119888
Also if 119891prime (1
2) = 0 them
61
4+ 2119887
1
2+ 119888 = 0
3
2+ 119887 + 119888 = 0
∵ 119888 = minus2
119887 =1
2
there4 2119887 + 119888 = 2(1
2) + (minus2)
= 1 minus 2
= minus1
6 If the points (1 1 120582) 119886119899119889 (minus3 0 1) are equidistant from the plane 3119909 + 4119910 minus 12119911 + 13 = 0
then 120582 satisfies the equation
1 31199092 + 10119909 + 7 = 0
2 31199092 + 10119909 minus 13 = 0
3 31199092 minus 10119909 + 7 = 0
4 31199092 minus 10119909 + 21 = 0
Answer (3)
Solution (1 1 120582) 119886119899119889 (minus3 0 1) in equidistant from 3119909 + 4119910 minus 12119911 + 13 = 0 then
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
|3 + 4 minus 12120582 + 13
radic32 + 42 + 122| = |
minus9 + 0 minus 12 + 13
radic32 + 42 + 122|
|20 minus 12120582| = |minus8|
|5 minus 3120582 | = |minus2|
25 minus 30120582 + 91205822 = 4
91205822 minus 30120582 + 21 = 0
31205822 minus 10120582 + 7 = 0
there4 Option 31199092 minus 10119909 + 7 = 0 Is correct
7 In a Δ119860119861119862119886
119887= 2 + radic3 119886119899119889 ang119862 = 60119900 Then the ordered pair (ang119860 ang119861) is equal to
1 (105119900 15119900)
2 (15119900 105119900)
3 (45119900 75119900) 4 (75119900 45119900)
Answer (1)
Solution Since 119886
119887=
2+ radic3
1 ang119860 gt ang119861
Hence only option 1 amp 4 could be correct checking for option (1) 119886
119887= sin105119900
sin 15119900
= 119904119894119899 (60119900 + 45119900)
sin(60119900 minus 45119900)= radic3 + 1
radic3 minus 1
119886
119887= 2 + radic3
1
Hence option (105119900 15119900) is correct
8 A factory is operating in two shifts day and night with 70 and 30 workers respectively If per
day mean wage of the day shift workers is Rs 54 and per day mean wage of all the workers is
Rs 60 then per day mean wage of the night shift workers (in Rs) is
1 75
2 74
3 69
4 66
Answer (2)
Solution 1198991 1199091 +1198992 1199092
1198991+1198992 = 119909
70 ∙ (54) + 30 (1199092)
70 + 30= 60
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
= 3780 + 30 1199092 = 6000
there4 1199092 = 6000 minus 3780
30
= 2220
30
= 74
9 The integral int119889119909
(119909+1)34 (119909minus2)
54
is equal to
1 4 (119909minus2
119909+1)
1
4+ 119862
2 minus4
3 (119909+1
119909minus2)
1
4+ 119862
3 4 (119909+1
119909minus2)
1
4+ 119862
4 minus4
3 (119909minus2
119909+1)
1
4+ 119862
Answer (2)
Solution int119889119909
(119909+1)34 (119909minus2)
54
Divide amp Multiply the denominator by (119909 + 1)5
4
int119889119909
(119909 + 1)2 (119909 minus 2119909 + 1
)
54
Put 119909minus2
119909+1= 119905
(1 (119909 + 1) minus (119909 minus 2)(1)
(119909 + 1)2) 119889119909 = 119889119905
3
(119909 + 1)2 119889119909 = 119889119905
1119889119909
(119909 + 1)2= 1
119889119905
3
rArr 13 int 1199055
4 119889119905 = 1 119905
14
3 (minus1
4)
= minus4
3 1
11990514
+ 119862
minus4
3 (119909+1
119909minus2)
1
4+ 119862
10 Let 119886 119886119899119889 be two unit vectors such that |119886 + | = radic3
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
If 119888 = 119886 + 2 (119886 times ) then 2|119888 | is equal to
1 radic51
2 radic37
3 radic43
4 radic55
Answer (4)
Solution As |119886 times | = radic3
Squaring both the sides
|119886 |2 + | |2+ 2119886 ∙ = 3
1 + 1 + 2 ∙ 1 ∙ 1 ∙ cos 120579 = 3
2119888119900119904120579 = 1
119888119900119904120579 =1
2
120579 = 60
there4 Angle between 119886 119886119899119889 119894119904 60119900
Now
|119888 | = |119886 + 2119887 + 3(119886 times 119887)|
Squaring both the sides
|119888 |2 = ||119886 |2 + 4| |2+ 9 (119886 times 119887)2 + 4 119886 ∙ (119887) + 3119886 ∙ (119886 times 119887) + 6119887 ∙ (119886 times 119887)|
|119888 |2 = |1 + 4 + 9 sin2 120579 + 4 119888119900119904120579 + 0 + 0 |
|119888 |2 = |5 + 93
4+ 4
1
2| =
55
4
there4 2|119888 | = radic55
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
119909 + 119910 + 119911 + 1 = 0 = 2119909 minus 119910 + 119911 + 3
Put 119911 = 1 so we get 119909 + 119910 + 2 = 0 and 2119909 minus 119910 + 4 = 0
We will get 119909 = minus2
119910 = 0
there4 The point (minus2 0 1) lies on the line and perpendicular vector will come from
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
|119894 119895 1198961 1 12 minus1 1
| = 2119894 + 119895 minus 3119896
So the equation line would be 119909 + 2
2=119910
1=119911 minus 1
minus3
And the other line 119909 minus 1
120572=119910 + 1
minus1=119911
1
Shortest distance would be
119863 = [(1198862 minus 1198861) 1198871 1198872]
|1198871 times 1198872|
When 1198861 = (minus2119894 + 119900119895 + 1119896)
1198862 = (119894 minus 119895 + 0119896)
1198871 = 2119894 + 119895 minus 3119896
1198872 = 120572119894 minus 119895 + 119896
|3 minus1 minus12 1 minus3120572 1 minus3
|
|119894 119895 1198962 1 minus3120572 minus1 1
|
= 3(1 minus 3) + 1 (2 + 3120572) + 1 (2 + 120572)
|minus2119894 minus 119895 (2 minus 3120572) + 119896 (minus2 minus 120572)|
|minus6 + 2 + 3120572 + 2 + 120572
radic4 + (2 + 3120572)2 + (2 + 120572)2| =
1
radic3
|4120572 minus 2|
radic4 + 4 + 12120572 + 91205722 + 4 + 4120572 + 1205722=
1
radic3
|4120572 minus 2
radic101205722 + 16120572 + 12| =
1
radic3
(161205722 minus 16120572 + 4)3 = 101205722 + 16120572 + 12
481205722 minus 48120572 + 12 =
101205722 + 16120572 + 12
381205722 minus 64120572 = 0
120572(19120572 minus 32) = 0
120572 =32
19
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
1+1199092) 119909 gt 1 then 119891(5) is equal to
1 120587
2
2 tanminus1 (65
156)
3 120587
4 4 tanminus1 (5)
Answer (3)
Solution
2 tanminus1 119909 + sinminus1 (2119909
1 + 1199092) 119891119900119903 119909 gt 1
= 2 tanminus1 119909 + 120587 minus 2 tanminus1 119909 119886119904 119909 gt 1
there4 119891(5) = 120587
there4 Answer is 120587
Or 119891(5) = 2 tanminus1 (5) + sinminus1 (10
26)
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
= 120587 minus tanminus1 (10
24) + tanminus1 (
10
24)
120587 sinminus1 (10
26)
29 Let the tangents drawn to the circle 1199092 + 1199102 = 16 from the point P(0 h) meet the 119909 minus 119886119909119894119904 at
points A and B If the area of Δ119860119875119861 is minimum then h is equal to
1 4radic2
2 3radic2
3 4radic3
4 3radic3
Answer (1)
Solution
Let the equation of the tangent be (119910 minus ℎ) = 119898 (119909 minus 0)
119898119909 minus 119910 + ℎ = 0
|ℓ119899
radic1198982 + 1| = 4
ℎ2 = 161198982 + 16
1198982 = ℎ2 minus 16
16
119898 = radicℎ2 minus 16
4
So co-ordinate of B would be
radicℎ2 minus 16
4 119909 minus 119910 + ℎ = 0
119909 = 4ℎ
radicℎ2 minus 16
Also of triangle
=1
2 119861119886119904119890 119909 119867119890119894119892ℎ119905
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Δ =1
2
8ℎ
radicℎ2 minus 16 ∙ ℎ
Δ = 4 ℎ2
radicℎ2 minus 16
119889Δ
119889ℎ= 4
[ 2ℎradicℎ2 minus 16 minus
2ℎ ∙ ℎ2
2radicℎ2 minus 16(ℎ2 minus 16)
]
= 4ℎ [4(ℎ2 minus 16) minus 2ℎ2
2radicℎ2 minus 16 (ℎ2 minus 16)]
=4ℎ[2ℎ2 minus 64]
2radicℎ2 minus 16 (ℎ2 minus 16)
For are to be minima ℎ = radic32
ℎ2 = 32
ℎ = 4radic2
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0