paper – 2 (physics) section 1: (only one option correct...
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Paper – 2 (PHYSICS) SECTION 1: (Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE Option is correct.
1. If Cu is the wavelength of K X-ray line of copper (atomic number 29) and Mo is the wavelength of the K X-
ray line of molybdenum (atomic number 42), then the ratio Cu/Mo is close to
(A) 1.99 (B) 2.14 (C) 0.50 (D) 0.48
Solution
(B) From Moseley’s law
Using 2
1
( 1)zl µ
- we get
2cu mo
2mo cu
( 1)
( 1)
Z
Z
l
l
-=
-Þ
241
2.1428
æ öç ÷ =ç ÷è ø
2. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds
of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1: u2 = 2:1 and
hc = 1240 eV nm, the work function of the metal is nearly
(A) 3.7 eV (B) 3.2 eV
(C) 2.8 eV (D) 2.5 eV
Solution
(A) From Einstein’s photoelectric equation, we know
20 max
1
2hv hv mv= + or 2
0 max
1
2
hcmvw
l= +
2max 0
1
2
hcmv w
l= -
21 0
1
1
2
hcmv w
l= - (1)
22 0
2
1
2
hcmv w
l= - (2)
Dividing Eq. (1) by Eq. (2), we get
Given 1
2
2u
u=
021 1
22
0
2
hc
u
hcu
wl
wl
-
=
-
Therefore,
0
10 0
2 10
2
4 44
1
hc
hc hc
hc
wl
w wl lw
l
-
= Þ - = -
-
34 8
19 19
6.6 10 10
1.6 10 10
shc
-
- -
´ ´ ´=
´ ´
hc = 1240 eV nm
Therefore,
0
4 1240 12403
310 248w
´- =
16 5 = 30 or 11 = 30 Þ 0 = 3.7 eV
3. Parallel rays of light of intensity I = 912 W m2 are incident on a spherical black body kept in surroundings of
temperature 300 K. Take Stefan–Boltzmann constant = 5.7 108 W m2 K4 and assume that the energy
exchange with the surroundings is only through radiation. The final steady state temperature of the black body is
close to
(A) 330 K (B) 660 K
(C) 990 K (D) 1550 K
Solution
(A) Effective area of incident light = rs2
Total energy per second = 912 r2
From Sebari’s law
4 40( )E Ae T Ts= -
For black body emissivity e = 1
A = 4r2
2912 rp´ 24 rp= 8 4 45.7 10 [ 300 ]T-´ ´ -
4 4 8
8
912( 300 ) 40 10
4 5.7 10T
-- = = ´
´ ´
T4 = 40 108 + 34 108
T4 = (90 + 81) 108 Þ 121 108
1/4(121) 100 11 100 3.3 100T = ´ = ´ Þ ´
T = 330 K
4. During an experiment with a metro bridge, the galvanometer shows a null point when the jockey is pressed at
40.0 cm using a standard resistance of 90 , as shown in the below figure. The least count of the scale used in the
meter bridge is 1 mm. The unknown resistance is
(A) 60 0.15 (B) 135 0.56
(C) 60 0.25 (D) 135 0.23
Solution
(C) In balanced condition
40 40
90 (100 4 ) 60
R
v= =
-
R = 60
In general
90 (100 )
R x
x=
-
log R = log x + log (100 x)
(100 )
dR dx dx
R x x=± ±
-
0.1 0.1 0.6 0.4 60
60 40 60 240
dRdR
+ ´=± ± Þ =
600.25
240= Þ W
(60 0.25)
5. A planet of radius 1
10R (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth
5
R on it and lower a wire of the same length and of linear mass density 103 kg m1 into it. If the wire is not
touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of
Earth = 6 106 m and the acceleration due to gravity of Earth is 10 m s2)
(A) 96 N (B) 108 N
(C) 120 N (D) 150 N
Solution
(B) As velocity will increase, centrifugal force will increase.
Radius of earth eR = 6 106 m
Radius of planet = 56 10 m
10eR
3
2 2
4 4
3 3e
e e
e
RGMg G GR
R R
p rp r
×= = × =
24 101 m s
3 10 10 10e
p
R gg Gp r -= × = = =
We know that 1d
g gR
æ ö¢ ç ÷= -ç ÷
è ø
Therefore, 1p
xg g
R
æ ö¢ ç ÷= -ç ÷
è ø at depth x
Mass of element of will of length dx
dm = dx
Force on element
1p
xdF dm g dx g
Rl
æ ö¢ ç ÷= × Þ × -ç ÷
è ø
1p
xdF g dx
Rl
æ öç ÷= -ç ÷è ø
/5 /5
0 0
1R R
p
xF g dx dx
Rl
é ùê ú= × -ê úë ûò ò
2
.5 25 2
p
R Rg
Rl
é ùê ú= -ê ú´ë û
(gp = 1 m s2)
3 510 9 6 10
50
- ´ ´ ´= = 108 N
6. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed
vertically on ground as shown in the below figure. The bead is released from near the top of the wire and it slides
along the wire without friction. As the bead moves from A to B, the force it applies on the wire is
(A) always radially outwards.
(B) always radially inwards.
(C) radially outwards initially and radially inwards later.
(D) radially inwards initially and radially outwards later.
Solution
(D) Initially the bead exerts a radially inwards force which contributes to the centripetal force. A stage comes when
N=0, bead loses contact with the wire as then re-establishes it at the lower end of the hole. Here N starts to act
radially outwards.
7. A glass capillary tube is of the shape of truncated cone with an apex angle so that its two ends have cross
sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its
cross section is b. If the surface tension of water is S, its density is , and its contact angle with glass is , the value
of h will be (g is the acceleration due to gravity)
(A) 2
cos( )S
b g
(B)
2cos( )
S
b g
(C) 2
cos( /2)S
b g
(D)
2cos( /2)
S
b g
Solution
(D)
From the figure
cos2
b Ra
qæ öç ÷= +ç ÷è ø
Excess pressure 2S
h gR
r=
or 2
cos2
sh g
b
ar q
æ öç ÷= +ç ÷è ø
2
cos2
sh
b g
aq
r
æ öç ÷= +ç ÷è ø
8. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and
2R, respectively, as shown in the below figure. If magnitudes of the electric fields at point P at a distance R from
the centre of spheres 1, 2 and 3 are E1, E2 and E3, respectively, then
(A) E1 > E2 > E3 (B) E3 > E1 > E2
(C) E2 > E1 > E3 (D) E3 > E2 > E1
Solution
(C) In first two cases P is outside sphere.
Therefore,
1 20
1
4
QE
E Rp= × (1)
2 20
1 2
4
QE
E Rp= × (2)
In case (iii) point P is inside
3
03E r
E
ræ öç ÷= ç ÷è ø
3
4
4(2 )
3
Q
R
r
p
=
3
12 3
4 8 8
Q Q
Rp p= =
×
3 3
3QE R
Rp
æ öç ÷= ×ç ÷è ø
3 20
1
4 2
QE
E Rp= × (3)
E2 > E1 > E3
9. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is
immersed in a lower refractive index liquid as shown in the below figure. It is found that the light emerging from
the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive
index of the liquid is
(A) 1.21 (B) 1.30
(C) 1.36 (D) 1.42
Solution
(C) In ∆SAB
11.54
2tan10
AB
ASq = =
tan = 0.577
c = 30
Using sin
,sin 90
c l
b
q m
m= we get
sin lc
b
mq
m= or
1sin 30 2.72
2a bm m= ´ = ´ = 1.36
10. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the
surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which
one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The
figures are only illustrative and not to be scale.
(A) (B)
(C) (D)
Solution
(B) When ball falls for time t, its velocity is v = u + at = 0 + gt and 21
2k mv=
2 21
2k mg t=
2k tµ
Graph is a parabola. During collision, it gets compressed and KE reduces till it becomes zero at maximum
compression.
SECTION 2: Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc. 6 questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answers among the four given options (A), (B), (C) and (D).
Paragraph for Questions 11 and 12: The figure shows a circular loop of radius a with two long parallel wires
(numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The
loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if
seen from above.
11. When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is
zero at a height h above the loop. In that case
(A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a
(B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a
(C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2a
(D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2a
Solution
(C) When current in two wires is in opposite direction the total magnetic induction directed into page is given by
B = B1 + B2
B = 2B1 (since 1 2B B=
)
0
2 2
2 2cos
4 ( )
IB
a b
mq
p= ×
+
0
2 2 2 2
2 2
4
I a
b h a b
m
p= × ×
+ +
0
2 2( )
IaB
a b
m
p=
+ (1)
Magnetic induction at P on axis of coil distant h from centre of drop of radius ‘a’.
20
2 2 3/2
2
4 ( )
IaB
a h
m p
p¢= ×
+
20
2 2 3/22( )
IaB
a h
m¢=+
(2)
If ,B B¢= from Eqs. (1) and (2), we get
20 0
2 2 2 2 3/2( ) 2( )
Ia Ia
a h a b
m m
p=
+ +
2 2 1/2
1
2( )
a
a hp=
+
2
2 2 2
4
( )
a
a bp=
+ 2 = 9.8596
2
2 2
4
9.8596
a
a h=
+
4h2 = 5.8596
5.85961.2
4h a a= @
12. Consider ,d a and the loop is rotated about its diameter parallel to the wires by 30 from the position shown
in the below figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new
position will be (assume that the net field due to the wires is constant over the loop)
(A) 2 2
0 I a
d
(B) 2 2
0
2
I a
d
(C) 2 2
03 I a
d
(D) 2 2
03
2
I a
d
Solution
(B) Total magnetic induction:
B = B1 + B2 Þ 2B1 Þ 0 22
4
I
d
m
p× ×
0IB
d
m
p= (1)
( )T NI A B= ´
T = N I AB sin (2)
N = 1; = 30; Area of loop
A = a2
Equation (2) becomes
Therefore, 2 0 1
2
IT I a
d
m
p= × ×
2 20
2
I aT
d
m=
Paragraph for Questions 13 and 14: In the figure a container is shown to have a movable (without friction) piston
on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between
outside and inside the container. The container is divided into two compartments by a rigid partition made of a
thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled
with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal
diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are 3
,2
VC R 5
,2
PC R and those
for an ideal diatomic gas are 5
,2
VC R 7
.2
PC R
13. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final
temperature of the gases will be
(A) 550 K (B) 525 K
(C) 513 K (D) 490 K
Solution
(D) T is final temperature, when equilibrium is reached
Heat gained = Heat lost
n2C
7 32 ( 400) 2 (700 )
2 2
R RT T´ ´ - = × × -
7T 2800 = 2100 3T
10T = 4900
T = 490 K
14. Now consider the partition to be free to move without friction so that the pressure of gases in both
compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
(A) 250 R (B) 200 R
(C) 100 R (D) 100 R
Solution
(D) Heat lost by mono atomic gas = Heat gained by diatomic gas
5 72 (700 ) 2 ( 400)
2 2R T R T´ ´ - = ´ -
3500 5T = 7T 2800
12T = 6300 or T = 525 K
Work done by mono atomic gas = n1R∆T (PV = nRT)
= 2 R (700 525) ∆T is negative
= 350R
Work done by diatomic gas = n2R∆T
= 2R (525 400)= 250 R. There is gain in temp.
Network done = 350 + 250 = 100R ∆T is positive
Paragraph for Questions 15 and 16: A spray gun is shown in the below figure where a piston pushes air out of a
nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid
container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is
sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm, respectively. The
upper end of the container is open to the atmosphere.
15. If the piston is pushed at a speed of 5 mm s1, the air comes out of the nozzle with a speed of
(A) a
l
(B) a l
(C) l
a
(D) l
Solution
(C) From equation of continuity
a1v1 = a2v2
(0.2)2 0.005 = (0.01)2 v2
400 0.005 = v2
v2 = 2 m s1
16. If the density of air is a and that of the liquid l, then for a given piston speed the rate (volume per unit time) at
which the liquid is sprayed will be proportional to
(A) a
l
(B) a l
(C) l
a
(D) l
Solution
(A) For horizontal flow, Bernoulli’s theorem becomes
21
2
pgh v k
r¢+ + =
21
2p v kr+ = (h = 0)
Between points 1 and 2
21 2
1
2a ap p vr- =
Between points 2 and 3
23 2
1
2e ep p vr- =
Since p1 = p3
Therefore,
2 21 1
2 2q e a av vr r=
e a
a e
v
v
r
r=
SECTION 3: Matching List Type (Only One Option Correct) This section contains 4 questions, each having two matching lists. Choice for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.
17. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at
rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the
following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift is
given in List II. Match the statements from List I with those in List II and select the correct answer using the code
given below the lists.
List I List II
P. Lift is accelerating vertically up. 1. d = 1.2 m
Q. Lift is accelerating vertically down 2. d > 1.2 m
with an acceleration less than the
gravitational acceleration.
R. Lift is moving vertically up with 3. d < 1.2 m
constant speed.
S. Lift is falling freely. 4. No water leaks out of the jar
Code:
(A) P-2, Q-3, R-2, S-4
(B) P-2, Q-3, R-1, S-4
(C) P-1, Q-1, R-1, S-4
(D) P-2, Q-3, R-1, S-1
Solution
(C) Horizontal range = Velocity of out flow Time taken
22
hH gh
g= ´
21
2h gt=
2ht
g=
2H h=
It is independent of value of acc. Due to gravity, it may be g, (g + a) or (g a).
For free fall ( ) 0g g g¢= - =
v = 0, so liquid will not flow
18. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = 2a, a, +a and +2a,
respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of
these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with
List II and select the correct answer using the code given below the lists.
List I List II
P. Q1, Q2, Q3, Q4 all positive 1. +x
Q. Q1, Q2 positive; Q3, Q4 negative 2. x
R. Q1, Q4 positive; Q2, Q3 negative 3. +y
S. Q1, Q3 positive; Q2, Q4 negative 4. y
Code:
(A) P-3, Q-1, R-4, S-2
(B) P-4, Q-2, R-3, S-1
(C) P-3, Q-1, R-2, S-4
(D) P-4, Q-2, R-1, S-3
Solution
(A)
1 4
2 3
F F
F F
=
=
Horizontal camps will cancel out. Net force will be along + y axis
Vertical camps of (F1 and F4); (F2 and F3) will cancel. Net force will be along + x-axis.
Horizontal camps of F1 and F4 will cancel as 1 4F F=
. Horizontal camps of F2 and F3 will cancel as 2 3F F=
vertical camp due to negative charges is more (distance being less).
Therefore, net force is along y-axis 1 4F F=
Vertical camps will cancel and horizontal camps get added along +x-axis
2 3F F=
Vertical camps get cancelled and horizontal camps are along – x-axis
Net force is along – x-axis.
19. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and
the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and
select the correct answer using the code given below the lists.<COMP: Please change numbering from P–S to (A)–
(D) and 1–4 to (P)–(S).>
Code:
(A) P-1, Q-2, R-3, S-4
(B) P-2, Q-4, R-3, S-1
(C) P-4, Q-1, R-2, S-3
(D) P-2, Q-1, R-3, S-4
Solution
(B) Using lens maker’s formula
1 2
1 1 1( 1)
f R Rm
æ öç ÷= - -ç ÷è ø
1 2
1 1 1
efff f f= +
Case P: 1 1 1
0.5f r r
æ öç ÷= -ç ÷
-è ø
r = f
Therefore, 1 1 1
efff r r= +
2
eff
rf =
P → (2)
Case Q: 1 1 1
0.5f r
æ öç ÷= -ç ÷
¥ -è ø
1
2r=
1 1 1
2 2efff r r= +
feff = r
Q → (4)
Case R: 1 1 1
0.5f r
æ öç ÷= -ç ÷
¥è ø
1
2r
-=
1 1 1
2 2efff r r= - -
feff = −r
R → (3)
Case S: 1 1 1 1
2 2efff r r r= - =
feff = 2r
S → (1)
20. A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see the below figure) on an inclined
plane with angle of inclination . Various values of are given in List I. The coefficient of friction between the
block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the
plane are equal to = 0.3. In List II expressions for the friction on the block m2 are given. Match the correct
expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration
due to gravity is denoted by g.
[Useful information: tan (5.5) ≈ 0.1; tan (11.5) ≈ 0.2; tan (16.5) ≈ 0.3]
List I List II
P. = 5 1. m2g sin
Q. = 10 2. (m1 + m2)g sin
R. = 15 3. m2g cos
S. = 20 4. (m1 + m2)g cos
Code:
(A) P-1, Q-1, R-1, S-3
(B) P-2, Q-2, R-2, S-3
(C) P-2, Q-2, R-2, S-4
(D) P-2, Q-2, R-2, S-3
Solution
(D)
Comp. of weight along AB
= m1g sin + m2g sin
= (m1 + m2)g sin (i)
Force of friction along BA
= 1m1g cos + 2m2g cos (1 = 0)
= 2m2g cos (ii)
(m1 + m2)g sin = 2mg cos
2
1 2
0.3 2tan 0.2
( ) 1 2
m
m m
mq
´= Þ Þ
+ +
Angle of sliding comes to be (11 30)
At 5 and 10, block will not slide and static friction is balancing (m1 + m2)g sin
For = 15 (>11.5) block will slide and friction is m2g cos .
Paper – 2 (CHEMISTRY) SECTION 1: (Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE Option is correct.
21. For the identification of -naphthol using dye test, it is necessary to use (A) Dichloromethane solution of -naphthol (B) Acidic solution of -naphthol (C) Neutral solution of -naphthol (D) Alkaline solution of -naphthol
Solution
(D) The reaction involved is
22. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is (A) Be2 (B) B2 (C) C2 (D) N2
Solution
(C) Out of the options, C2 has MO configuration 2 2 2 2 2 1 11 *1 2 *2 2 2 2z x ys s s s p p p , with two unpaired
electrons, so it is paramagnetic.
23. For the elementary reaction M N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is (Chapter chemical kinetics) (A) 4 (B) 3 (C) 2 (D) 1 Solution (B) Rate = k[M]n. If n = 3, then on increasing the concentration of M two times we get
Rate = k[2M]3 = 8k[M]3
24. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.
The correct order of their boiling point is (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II Solution (B) As the branching increases, the boiling point decreases. This is because the van der Waals forces of
attraction decrease.
25. The acidic hydrolysis of ether (X) shown below is fastest when
(A) One phenyl group is replaced by a methyl group (B) One phenyl group is replaced by a para-methoxyphenyl group (C) Two phenyl groups are replaced by two para-methoxyphenyl groups (D) No structural change is made to X Solution (C) This is because –OCH3 is an electron donating group which increases the stability of the carbocation.
26. The major product in the following reaction is
Solution (D) The reaction is
27. Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a (A) Reducing agent, oxidizing agent (B) Reducing agent, reducing agent (C) Oxidizing agent, oxidizing agent (D) Oxidizing agent, reducing agent Solution (A) The reactions involved are
H2O2 + KIO4 KIO3 + H2O + O2
H2O2 + NH2OH H2O + N2O3
28. The product formed in the reaction of SOCl2 with white phosphorus is (A) PCl3 (B) SO2Cl2 (C) SCl2 (D) POCl3 Solution (A) The reaction involved is
P4 + 8SOCl2 4PCl3 + 4SO2 + 2S2Cl2
29. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is
(A) 0 (B) 1 (C) 2 (D) 3 Solution (C) The reaction involved is
XeF6 + H2O (complete hydrolysis) XeO3 + H2F2 2OH /H O
4HXeO XeO6 + Xe (g) + H2O + O2(g)
30. For the process H2O(I) H2O(g) at T = 100C and 1 atmosphere pressure, the correct choice is (A) Ssystem > 0 and Ssurroundings > 0 (B) Ssystem > 0 and Ssurroundings< 0 (C) Ssystem < 0 and Ssurroundings > 0 (D) Ssystem < 0 and Ssurroundings < 0 Solution (B) We know that ΔSsystem + ΔSsurroundings = 0 ΔSsystem = ΔSsurroundings. Therefore, ΔSsystem > 0 and ΔSsurroundings
< 0.
SECTION 2: Comprehension Type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiments, data etc. 6 questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answers among the four given options (A), (B), (C) and (D).
Paragraph for Questions 31 and 32: X and Y are two volatile liquids with molar weights of 10 g mol1 and 40 g mol1, respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapors of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behavior for the inert gas and the two vapors. (Chapter gaseous state)
31. The value of d in cm (shown in the figure), as estimated from Graham's law, is (A) 8 (B) 12 (C) 16 (D) 20 Solution (C) Using Graham’s law of diffusion we have
1/240
1624 10
xx
x
32. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to (A) Larger mean free path for X as compared to that of Y (B) Larger mean free path for Y as compared to that of X (C) Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas (D) Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas Solution
(D) Collision frequency of X is greater than Y.
Paragraph for Questions 33 and 34: Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.
33. The product X is
Solution (A) The reactions involved are
34. The correct statement with respect to product Y is (A) It gives a positive Tollens test and is a functional isomer of X (B) It gives a positive Tollens test and is a geometrical isomer of X (C) It gives a positive iodoform test and is a functional isomer of X (D) It gives a positive iodoform test and is a geometrical isomer of X Solution
(C) As it has CH3CO– group so it gives positive iodoform test and is an isomer of X (only the functional
groups are different).
Paragraph for Questions 35 and 36: An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below: Scheme:
35. M1, Q and R, respectively are (A) Zn2+, KCN and HCl (B) Ni2+, HCl and KCN (C) Cd2+, KCN and HCl (D) Co2+, HCl and KCN Solution (B) The reactions involved are
Ni2+ + HCl [NiCl4]2
Ni2+ + KCN [Ni(CN)4]2
36. Reagent S is (A) K4[Fe(CN)6] (B) Na2HPO4 (C) K2CrO4 (D) KOH Solution (D) The reaction involved is
Zn2+ + OH Zn(OH)2 OH
[Zn(OH)4]2 (soluble)
SECTION 3: Matching List Type (Only One Option Correct)
This section contains 4 questions, each having two matching lists. Choice for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.
37. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists. {en = H2NCH2CH2NH2; atomic numbers : Ti = 22, Cr = 24; Co = 27; Pt = 78}
Code: P Q R S (A) 4 2 3 1 (B) 3 1 4 2 (C) 2 1 3 4 (D) 1 3 4 2 Solution
(B) For [Cr(NH3)4Cl2]Cl, Cr (+1) = 3d5
For [Ti(H2O)5Cl](NO3)2, NO3 is an ambidentate ligand Ti (+3) = 3d5
For [Pt(en)(NH3)Cl]NO3, NO3 is an ambidentate ligand Pt (+2) = 5d8
For [Co(NH3)4(NO3)2]NO3, NH3 is a strong ligand and causes pairing Co (+3) = 3d6
38. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists.
Code: P Q R S (A) 2 1 3 4 (B) 4 3 1 2 (C) 2 3 1 4 (D) 4 1 3 2 Solution (C) Concept based.
39. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists.
Code: P Q R S (A) 1 3 4 2 (B) 2 4 3 1 (C) 4 1 2 3 (D) 3 2 1 4 Solution (A) The reactions are as follows:
40. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.
Code: P Q R S (A) 1 4 2 3 (B) 3 1 4 2 (C) 3 4 2 1 (D) 4 1 3 2 Solution (C) The reactions involved are as follows:
Paper-2 (MATHEMATICS)
SECTION 1 : (One or More Than One Options Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct.
41. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that eachenvelope contains exactly one card and no card is placed in the envelope bearing the same number andmoreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it canbe done is (A) 264 (B) 265 (C) 53 (D) 67 Solution (C)
1 2 3 4 5 6
1 2 3 4 5 6
]
This is problem of derangement.
Number of derangements of n things is 0
( 1) 1 1 1 ( 1)1
1 2 3
k nn
k
n nk n
L
In the equation, two possibilities are there I When card two goes to envelope 1
Derangement of 3, 4, 5, 6 cards is 4 11
1
1
2
1
3
1
4
1 1 124 12 4 1
2 6 24
9 ways.
II. When card two does not go to envelope one.
Now it is derangement of 2, 3, 4, 5, 6 in envelopes 1 3 4 5 6 in 1 1 1 1 1
5 11 2 3 4 5
120 11
1
1 1 1 160 20 5 1
2 6 24 120
44 ways.
Total number of way = 9 + 44 = 53 42. In a tringle the sum of two sides is x and the product of the same two sides is y. If x2 − c2 = y, where c is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is
(A) 3
2 ( )
y
x x c
(B) 3
2 ( )
y
c x c
(C) 3
4 ( )
y
y x c
(D) 3
4 ( )
y
c x c
Solution (B) Let a + b = x (1) ab = y (2) x2 – c2 = y (3)
From (1) a + b + c = c + x (4) From (2) abc = cy (5)
(4) 2
2 2s c x c x rr c x
rs
Q
(5) 44 4
cy abcR cy R R
Q
22 4 8
( )
r
R c x abc c x abc
2
2
18 sin
22
( )( )
ab Ca
c x abc
2b 2sin
( )
C
c x a bc (6) Now from (3) (a + b)2 – c2 = y or 2abcosc + 2ab = y
or 2 ab (1 cos )c ab
1cos2
c
Now from (6) 2
12 1
(1 cos ) 4
( ) ( )
abr ab c
R c x c c x c
32
3 34( ) 2 ( ) 2 ( )
abab y
c c x x x c c x c
43. The common tangents to the circle x2 + y2 =2 and the parabola y2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area quadrilateral PQRS is (A) 3 (B) 6 (C) 9 (D) 15 Solution (D)
.
We know eqn. of tangent to parabola with slope m is 2
y mxm
(1)
Since (1) is also tangent to the circle, so length of from centre is equal to the radius.
2 2
2(0) 0
2( 1)
mm
m
22
4 2( 1)mm
or
4 = 2m2 (m2 + 1) or 2 = m4 + m2
or 2 2 2 2 1 1 8( ) 2 0
2m m m
2 2 21 31, 2 2 1
2m m m
Q
1m
Tangents are 2
1y x and
2
1y x
Pts of contact 2
2 4,
1 1
2
2 4,
( 1) 1
(Parabola)
i.e. (2, 4) and (2, 4) )(S and R)
2
2,
a a
m m
Q for a
y mxm
For pt. of contact of circle :- Solving y = x + 2 and x2 + y2 = 2 x2 + (x + 2)2 x2 + x2 4 + 4x – 2 =0 2x2 +4x + 2 = 0 x2 2x + 1 =0 (x + 1)2 = 0 x = – 1 y = 1 (–1, 1) and (–1, –1) By symmetry are P and Q.
Now area of the trapezium PQRS is 1
( ) Distance2
PQ RS
i.e. 1
{(2 8)(1 2)}2
1
2 10
53 15
44. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more then the number of girls ahead of her, is
(A) 1
2
(B) 1
3
(C) 2
3
(D) 3
4
Solution (A) According to question following possibilities are there Case 1 G1 B1 G2 B2 B3 B1 G1 B2 G2 B3 G1 B1 B2 G2 B3 Girls separate
1 2 3B B B
Out of 3 gaps, 2 are selected and girls are standing there in 32C ways.
Next Boys and girls permute separately in 3 2 ways.
Number of ways 32 3 2
3 6 2 36
C
Case 2 G1 G2 B1 B2 B3 B1 G1 G2 B2 B3 B1 B2 G1 G2 B3 Not possible Girls together
Places selected in 21 2C ways (Gaps) and then permutation is 3 2 ways.
Number 21 3 2 2 6 2 24C ways.
Probability 36 24 60
5
120
1
22
45. The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (A) only purely imaginary roots (B) all real roots (C) two real and two purely imaginary roots (D) neither real nor purely imaginary roots Solution (D) Given equation is p(x) = 0 Let it be written as x2 + c = 0 where c > 0 ( purely imaginary roots) p(p(x)) = (p(x))2 + c = 0 or (x2 + c)2 + c = 0 or x4 + c2 +2cx2 + c = 0 or (x2)2 + 2cx2 + c2 + c = 0
2 2
2
2
2 4 4( )
2
2 2
x c c cx
c c
2cwhere 0
2
cc ic c
x c ic Hence p(p(x)) = 0 has neither real nor purely imaginary roots. Note:– Let c = 1 Let us find square root of –1 + 2
Let 2 1 2 2 cos 2 sin4 4
x
14
1
23 32 cos 2 2 sin 2
4 4x k k
k = 0, 1
1 14 4
3 3 11 112 cos 2 sin , 2 cos 2 sin
4 4 4 4
46. For x (0, ), the equation sin x + 2sin2x −sin3x = 3 has (A) Infinitely many solutions (B) Three solutions (C) One solution (D) No solution Solution (D) Given equation is sin x + 2sin x – sin 3x = 3
or 3sin 4sin cos 3sin 4sin 3x x x x x
or 2sin [1 4cos 3 4sin ] 3x x x
or 21 4cos 3 4(1 cos ) 3cosx x ecx
or 22 4cos 4cos 3cosx x ecx
or 2 24 cos cosx x
43cos
2ecx
or 2 1 1 14 cos cos 3cos
4 4 2x x ecx
or 2
1 34 cos 3cos
2 4x ecx
min
11min 0at cos
2
3max at
2
x
min = 3 at 2
3max ( 4) 3
4
at 1 1
cos2 2
x
min ( 4 6
)4
6 at
Through max of LHS = min of RHS but max of LHS occurs at a different point i.e. 1 1cos
2 2
from min of RHS
which occurs at 2
.
Within (0, ) rather in the whole domain, graph never meet. Elaboration through graphs
Note:– (1) If max of LHS > and min of RHS < even then we cannot decided that there is a solution.
47. The following integral 2
17
4
(2cosec )x dx
is equal to
(A) log(1 2 )
16
02( )u ue e du
(B) log(1 2 )
17
02( )u ue e du
(C) log(1 2 )
17
02( )u ue e du
(D) log(1 2 )
16
02( )u ue e du
Solution
(A) 172
4
(2cosec )x dx
Let cosec cot tx x e
2( cosec cot cosec ) tx x x dx e dt
cosec (cot cosec ) tx x x dx e dt
(cosec ) t tx e dx e dt
(cosec )x dx dt
Now when 4
x
,
cosec cot4 4
te
2 1te
log (1 2)et
when 2
x
, cosec cot2 2
te
1 – 0 = et t = 0 Also cosec2x – cot2x = 1
1
cosec cot t
tx x e
e
2 cosecx = et + e–t Integral reduces to
0 log (1 2 )16
log (1 2) 0( ) 2 2 ( )
e
e
t t t te e dt e e dt
Note: t can be replaced by . 48. Coefficient of x11 in the expansion pf (1 + x2)4 (1 + x3)7 (1 + x4)12 is (A) 1051 (B) 1106 (C) 1113 (D) 1120 Solution (C) Expanding by Binomial
Coeff. of 11x in 4 2 0 4 2 1 4 2 2 4 2 3 4 2 4
0 1 2 3 4
7 3 0 7 3 1 7 3 2 7 3 3 7 3 40 1 2 3 4
{ ( ) ( ) ( ) ( ) ( )( ) }
{ ( ) ( ) ( )( ) ( )( ) ( )( ) }
C x C x C x C x C x
C x C x C x C x C x
further not required. 12 4 0 12 4 1 12 4 2
0 1 2{ ( ) ( ) ( ) }C x C x C x
Coeff. of 11x in 2 4 6 8
3 6 9
4 8
{1 4 6 4 }
{1 7 21 35 }
{1 12 64 }
x x x x
x x x
x x
Coeff. of 11x in 3 6 9 2 5 8 11 4 7 10 6 9 8 11 4 8{1 7 21 35 4 28 84 140 6 42 126 4 28 7 } {1 12 66 }x x x x x r x x x x x x x x x x 3 6 9 2 5 8 11 4 7 10 6 9 8 11 4 8{1 7 21 35 4 28 84 140 6 42 126 4 28 7 } {1 12 66 }
462 140 504 7
1113
x x x x x r x x x x x x x x x x
Option C is correct. 49. Let f : [0, 2] → be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1.
Let F (x)
2
0
( ) ( )x
F x f t dt for x [0, 2]. If F’(x) = f’ (x) for all x (0, 2) then F (2) equals
(A) e2 − 1 (B) e4 − 1 (C) e − 1 (D) e4 Solution (B)
2
0( )
x
f x f d dt
2 2( ) ( ) 0 0d d
f x f x x fdx dx
(By Newton-Leibnitz rule)
22xf x (1)
Now according to question
2( ) ( ) 2 ( )f x f x xf x f x
or 2 ( ) ( )xf x f x (0, 2)x
( )
2( )
f xx
f x
( )2
( )
f xdx xdx
f x
2
log ( )e f x 2
2
xc
2 2
( ) x c x cf x e e e
By initial condition f(0) = 1
2
1 o ce e ec = 1
2
( ) xf x e 2
t tf t e e
Now 2
2 2
00
( ) [ ] 1x
t t x xf x e dt e e
22 4(2) 1 1f e e
50. The function y = f (x) is the solution of the differential equation 4
2 2
2
1 1
dy xy x x
dx x x
in (−1.1) satisfying f (0)
= 0. Then 3
2
( )f x dx is
(A) 3
3 2
(B) 3
3 4
(C) 3
6 4
(D) 3
6 2
Solution
(B) D.E. is 4
2 2
2
1 1
dy xy x x
dx x x
or 4
2 2
2
1 1
dy x x xy
dx x x
(1)
It is a linear differential eqn.
2
21
log(1 )1 2
2
I.F.
log 1
xdx x
x
e
e e
e x
Now multiplying I.F. with (1)
2 2 4
21 1 2
(1 )
dy xx x y x x
dx x
2 4( 1 ) 2d
x y x xdx
2 4( 1 ) ( 2 )d x y x x dx
5 2
2 21
5 2
x xx y c (2)
Using initial conditions 21 (0) 0 0x c c = 0
(2) given 5
2 215
xx y x
52
5 2
2 2 2
5
1 5 1 1
xx
x xy
x x x
Now
3 3 35 2
2 2 23 3 32 2
2 2 2
( )5 1 1
x xf x dx dx dx
x x
3 2
2
200 2
1
xdx
x
(3)
odd function even function
Now solving 3 2
2
20 1
xdx
x
Putting x = sin dx = cos d
3 2
2 3
0 0
sin cos 1 cos 2
cos 2
dd
3 3
0 0
1 1 sin 2
2 2 2
1 1 2sin
2 3 4 3
1 3 3
6 4 2 6 8
From (3)
Answer is 3 3
26 8 3 4
.
SECTION 2: Comprehension Type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D).
Paragraph for Questions 51 and 52: Box 1 contains three cards bearing numbers 1, 2, 3, box 2 contains five five bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be number of the card drawn from the ith box, i = 1,2,3.
51. The probability that 1 2 3x x x is odd, is
(A) 29
105
(B) 53
105
(C) 57
105
(D) 1
2
Solution
(B) | (1)(2)(3) | | (1)(2)(3)(4)(5) | | (1)(2)(3)(4)(5)(6)(7) |
1 2 3
For x1 + x2 + x3 odd either all the number are odd 2 × 3 × 3 × 4 =24 ways or one odd and two even in
odd from 3rdodd from Ist odd from second
2 2 3 1 3 3 1 2 4 14 2 4314 2 43 1 2 3
= 12 + 9 + 8 = 29 ways Total number of ways = 24 + 29 = 53 ways. All possibilities = 3 × 5 × 7 = 105
Probability 53
105
52. The probability that x1, x2, x3 are in an arithmetic progression, is
(A) 9
105
(B) 10
105
(C) 11
105
(D) 7
105
Solution (C) For x1, x2, x3 in A.P. 2x2 = x1 + x3 we require x1 + x3 even either both even in = 1 × 3 =3 ways or both x1 , x3 odd is 2 × 4 = 8 ways All possibilities = 3 × 5 ×7 = 105
Probability Favorable ways
All possibilities
8 3 11
105 105
Paragraph for Questions 53 and 54: Let a, r, s, t be nonzero real numbers, Let P (at2 2at).Q.R (ar2. 2ar) and S (as2.2as) be distinct points on the parabola y2 = 1ax. Suppose that PQ is the focal chord and lines QR and PK are parallel. Where K is the point (2a, 0). 53. The value r is
(A) 1
t
(B) 2 1t
t
(C) 1
t
(D) 2 1t
t
Solution (D) Since QR || PK slope of QR = Slope of PK
or 2
2
2
22
2
22
aar
atta at art
or 2
2
2
1
1 2
rtt
trt
or
1r
t
1r
t
2 21
t
tr
t
or tr − 1= t2 − 2
2 1t
rt
Q PQ is a focal chord
1 1
11t t t
t
21 1(Q( ,2 ))at at
2
2Q ,
a a
t t
54. If st = 1. then the tangent at P and the normal at S to the parabola meet at a point ehose ordinate is
(A) 2 2
3
( 1)
2
t
t
(B) 2 2
3
( 1)
2
a t
t
(C) 2 2
3
( 1)a t
t
(D) 2 2
3
( 2)a t
t
Solution
(B) Q st = 1 1
st
2
1 2,
aS a
t t
2
2
2,
a aS
t t
Now eqn. of tangent of P is yt = x + at2 (1) Eqn. of normal at S is y = − x + 2as + as3
3
2x a a
t t t (2)
From (1) x = yt − at2
From (2) 3
2a ax y t
t t
2
22
ayt at a yt
t
2 2
2 2
2
12 2 2
1
ayt at a a t
t t
a tt
2
2
3
1
( 1)
2 2
a ta tt
yt t
Option B is correct.
Paragraph for Questions 55 and 56: Given that for each a (0, 1). 1
1
0lim (1 )
ha a
hh
t t dt
exixts. Let this limit be
g(a). In addtion. It is given that the function g(a) is differentiable on (0, 1).
55. The value of 1
2g
is
(A) π (B) 2 π
(C) 2
(D) 4
Solution
(A) Given 1
1
0( ) lim (1 )
ha a
hhg a t t dt
0
1 11 1
2 2
0
1 1lim (1 )
2 (1 )h
h h
h hhg t t dt dt
t t
1 1
0 02
2
1 1
1 1
4 4
h h
h hh h
dt dtt t
t t
1 1
0 2 2 20
1 1
1 1 1 1
2 4 2 2
h h
h h hhdt dt
t t
1
11 10 0
1
2sin sin (2 1)1
2
h
h
h h h
h
tt
1 10 [sin {2 2 1} sin (2 1)]h h h
1 1
0 0sin (1 2 ) sin (2 1)h hh h
2 2
56. The value of 1
'2
g
is
(A) 2
(B) π
(C) 2
(D) 0 Solution
(D) 1
1
0( ) lim (1 )
hQ Q
hhg a t t dt
(1)
Now 1
(1 ) 1 1
0(1 ) lim (1 )
ha a
hhg a t t dt
11
0lim (1 )
ha a
hht t dt
0lim (h
h
1 h 1
1) {1 (h
a
ht h
1 h )} at dt
10 (1 ) (1a
h t 1
1
)h
a
ht dt
11
0 (1 )h
a ah h
t t dt
(2)
From (1) & (2) g (a) = g (1 − a) g’ (a) = − g’ (1 − a)
when 1
2a ,
1 1' '
2 2g g
1
' 02
g
SECTION 3: Matching List Type (Only One Option Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct. 57. List-I List-II P. The number of polynomials f(x) with non-negative integer
coefficients of degree 2, satisfying f(0) = 0 and 1
0( ) 1,f x dx is
1. 8
Q. The number of points in the interval [ 13, 13] at which f(x) =
sin(x2) + cos(x2) attains its maximum value, is
2. 2
R. 22
2
3
(1 )x
xdx
e equals 3. 4
S. 1
21
2
1
2
0
1cos 2 log
1
1cos2 log
1
xx dx
x
xx dx
x
equals
4. 0
P Q R S (A) 3 2 4 1 (B) 2 3 4 1 (C) 3 2 1 4 (D) 2 3 1 4 Solution (D) P. Let f (x) = ax2 + bx f (0) = 0
Also 1 1
2
0 01 ( ) 1fxdx ax bx dx
13 2
0
1 13 2 3 2
ax bx a b
0, 0a b Q . and both are integration
0 2
13 2 a = 3, b = 0
or 3 0
1 3, 03 2
a b Q
Possible polynomials are f(x) = bx or f(x) = 3x2 P → 2
. 2 2 2 21 1( ) sin cos 2 sin cos
2 2f x x x x x
22 sin4
x
Now f(x) is maximum when 2
4 2x
Generalizing for whole domain
2 24 2
x k
, where k is integer and 2 is period.
2 24
x k
Q ≈ 3.14
when k = 0, 2
4x
[0, 13]
when k = 1, 2 9
4x
[0, 13]
S. Let 1
( ) cos 21
xf x xfx
x
1
( ) cos21
xf x xfx
x
1
( ) ( ) cos 2 n
xf x f x x f
1 x
1 x
1 x
0
f(x) is odd function
12
12
1cos 2 0
1
xx fx dx
x
S 4 Denominator will be non-zero. 58. List-I List-II P.
Let y(x) −cos (3 cos −1 x), x[−1, 1], x 3
2.Then
22
2
1 ( ) ( )( 1)
( )
d y x dy xx x
y x dx dx
equals
1. 1
Q. Let 1, T .n (n > 2) be the vertices of a regular polygon of n sides with its centre at the origin. Let ak be the position vector of the point k
1 11 1 1 11.2. . .If ( ) ( ) .n n
k k k k k kk n a a a a then the minimum value
of n is
2. 2
R. If the normal from the point P(h.1) on the ellipse
2 2
16 3
x y is perpendicular
to the line x + y − 8. then the value of h is
3. 8
S. Number of positive solutions satisfying the equation
] ] ]
2
1 1 2tan tan tan
2 1 4 1x x x
is
4. 9
P Q R S (A) 1 3 2 1 (B) 2 4 3 1 (C) 4 3 1 2 (D) 2 4 1 3 Solution
(A) 1P ( ) cos(3cos )y x x
1
1
2 2
1 3sin(3cos ){sin(2cos )}3
1 1
dy xx
dx x x
2
22 11 2sin(3cos )dy
x xdx
or 2
2 2 1(1 ) 9sin (3cos )dy
x xdx
2 1
2
9{1 cos (3cos )}
9{1 }
x
y
None differentiating 22
2
2(1 )2 ( 2 ) 9 2
dy d y dy dyx x y
dx dx dx dx
or 2
2
2(1 ) 9
d y dyx x y
dx dx
or 2
2
2( 1) 9 4.
d y dyx x y P
dx dx
When k = 2, 2 7
4x
[0,13]
There are only two values of x2 and hence 4 values of x. Q 3
R. 2 0 22 2 2
2 2 0
3 3 3
1 1 1x x x
x x xdx dx dx
e e e
(1)
Now 0 02 2
2 2
3 3( )
1 1x t
x tdx dt
e e
Putting x = t x = –2 t = 2 x = 0 t = 0
0 2
2
3
1 t
tdt
e
2 2
0
3
1 x
xdx
e
Now from (1) 2 2 22 2 2
2 0 0
3 3 3
1 1 1x x x
x x xdx dx dx
e e e
2 22 2
10 0
1 1 13 3
1 1 1 1x
x
x x x
dxe
ex dx x
e e e
2 1(3 )
xex
1xe
22 3
0 0
3
33
2 0 8 1
xdx
R
(Q) 1
1 1 2 2 3 11
( ) | |n
k k n nk
a a a a a a a a
r r r r r r r r
L
2 2 2 2( )sinn
a a a
2 2( 1) sinn
n a (1)
Q all 1| |a
equal
Also 1
2 2 21
1
2( ) ( ) cos
n
k kk
a a a a an
2 2( 1) cosn a
n
(2)
From (1) & (2) 2 2
sin cosn n
2 2
tan 1n
n
8
4n
Q 3
(R) Eqn of normal at ( 6 cos , 3 sin ) is
6 3
3cos sin
x y
(1) 2 2
cos sin
ax bya b
Slope of this normal is
6 sin6 32 tan
cos sin cos3
(h, 1) lies an this normal
6 3
3cos sin
h l
(2)
Now this normal is to x + y = 8 its slope in 1
11
2 tan = 1 tan = 1
2
2 1
cos , sin3 3
(2) 12
33
6 33
h
3 6 3
62
h
26
3h h = 2
R 2
S 1 1 1
2
1 1 2tan tan tan
2 1 4 1x x x
1 1
2
1 132 1 4 1tan tan
11
(2 1)(4 1)
x xx
x x
2
1 1
2 2
2
4 1 2 128 6 1tan tan
8 6 1 1
8 6 1
x x
x xx x x
x x
1 1
2 2
6 2 2tan tan
8 6
x
x x x
3 2 26 2 16 12x x x x
22 3 8 6 0x x x x
2(3 7 6) 0x x x
23 9 25 6 0x x x
3 ( 3) 2( 3) 0x x x
or ( 3)(3 2) 0x x x
x = 0, 3, 2
3
Number of +ve solutions = 1 S 1 59. Let f1: , f2: [0, ) , f11: and f4: [0, ) be defined by
1
22
3
| | if 0.( )
if 0 :
( )
sin if 0.( )
if 0
x
x xf x
e x
f x x
x xf x
x x
and
2 1
4
2 1
( ( )) if 0( )
( ( )) 1 if 0
f f x xf x
f f x x
List-I List-II P. f4 is 1. Onto but not one-one Q. f4 is 2. Neither continuous nor one-one R. f2 of1 is 3. Differentiable but not one-one S. f2 is 4. Continuous and one-one P Q R S (A) 3 1 4 2 (B) 1 3 4 2 (C) 3 1 2 4 (D) 1 3 2 4 Solution (D)
2
22 1 1
22
| | , 0 & | | valid
for , ( )
( ( ) ( ( ))
Range Dom ( ) , 0 & validx x
x x
f x R
f f x f x
f f e e
22 1
2
( ( )) 0
0x
f f x x
e
2
2
, 0
( )
1, 0n
x
x x
f x
e x
(P) f 4 Dom. R Range [0 ∞] Co-dom. = [0 ∞] onto
Now LHL at 0 = 2
0lim 0x
x
RHL at 0 = 2
0lim 1 1 1 0x
xe
f4 (0) = 0 Continuous at 0.
Not LHD at 2
0 0 0
(0 ) (0) 00 0
h h h
f h f hh
h h
RHD at 0 2
0 0
(0 ) (0) 1 0h
h h
f h f e
h h
2
0
1.2 1 2 2
2
h
h
e
h
Not derivable at 0 Not one-one (obvious from graph) P 1 Q. f3 in neither one-one nor onto. Derivability at 0
LHD = 0 0
sin(0 ) 0 sinh1h h
h
h h
RHD = 0
0 01h
h
h
Derivable at 0 Q 3 R 2 (obvious from graph). S 4 (obvious from graph.)
60. Let 2 2
cos sin ; 1,2, ,9.10 10
k
k kz i k
List-I List-II P. For each zk there exists a zj such that zk . zj =
1
1. True
Q. There exists a k {1, 2, , 9} such that z1.z = zk has no silution z in the set of complex numbers
2. False
R. 1 2 9|1 ||1 | |1 |
10
z z z equals
3. 1
S. 9
11 cos10
k
k
equals
4. 2
P Q R S (A) 1 2 4 3 (B) 2 1 3 4 (C) 1 2 3 4 (D) 2 1 4 3 Solution
(C) P 2 2
cos sin , 1,2, 910 10
k
k kz i k
L
Note 2 2(10 ) 2
( 10 )10 10 10
10
i k i k ik k
k kz z e e e
2
1010 cos 2 sin 2 1 (0)
1
i
e i i
10 1k kZ Z
10
1 9
2 8
9 1
1
1
1
1 1
k kz z
z z
z z
z z P
M
Q.
210 2( 1)
101 1
21 10
ki
i kk
k ki
z ez z z z e z
ze
for k = 1, 0 cos0 sin 0 1z z i
for k = 2, z = z1
for k = 9, z = z8
Solutions are these Q 2 R We know if 1, Z1 Z2, Zn are n, nth roots of unity them they are Roots of Zn – 1 = 0
1 2 1( 1) ( ) ( ) ( ) ( 1)nnz z z z z z z z L
1 2( 1)( ' 1)n nz z z z L
1 21 2 1( ) ( ) ( ) 2 ' 1n n
nz z z z z z z z L L
Putting z = 1
1 2 1(1 )(1 ) (1 ) 1 1 1nz z z n L L
1 2 9(1 )(1 ) (1 ) 10z z z L
1 2 91 1 1 10z z z L
or 1 2 91 1 11
10
z z z
L
R 3
S We know 2 11 0n L sum of n, nth Roots of 1
0 1 2 9 0z z z z L
1 2 9 01 { } 1 ( ) 1 ( 1)
2
z z z z
L
S 4