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SOLUTIONS TO IRP FULL TEST-2 ONLINE (HELD ON 12TH & 14TH MAY 2018) PHYSICS (PAPER-1)
SECTION 2:
1. In the
figure two
cylindrical conductor
s with same cross-sections and different
resistivity’s A and B are put end to end. A
current I is flowing from A to B. C is the
boundary of conductor A and B. VA and VB
are potential difference across the conductor
respectively. Select the correct option/s
(A) EA < EB if A B
(B) EA = EB if A B
(C) At the common boundary some charge will
be developed
(D) VA = VB if A B and lA = lB
Solution (ABCD):
E
J E
Also if A B A B A B A B& R R V V
2. Two capacitors C1
and C2 are charged
to q1 and q2 and
connected with
capacitor C as shown in figure. As switch S is
closed:
(A) C gets extra charge when 21
1 2
CC
q q
(B) C gets extra charge when 21
1 2
CC
q q
(C) C gets extra charge when 21
1 2
CC
q q
(D) C does not gets extra charge for any value
of q1 and q2.
Solution (AB): C will not get any extra charge only if voltage
across C1 and C2 is same
3. The diagram shows a
circuit with two identical
resistors. The battery has a negligible internal
resistance. When switch
S is closed, (A) Equivalent resistance of the circuit
decreases (B) Ammeter reading will increase
(C) Voltmeter reading will increase
(D) Power dissipated across R in right branch
will become zero.
Solution (ABCD):
One of the resistor gets shunted if key is
closed so current in circuit will increase.
4. A thin conducting rod of length l is moved
such that its end B
moves along the X-axis
while end A moves
along the Y-axis. A
uniform magnetic field 0ˆB B k exists in the
region. At some instant, velocity of end B is v
and the rod makes an angle of 60 with
the x-axis as shown in the figure. Then, at
this instant
(A) angular speed of rod AB is 2v
3
(B) angular speed of rod AB is 3v
2
(C) e.m.f. induced in rod AB is B v 3
(D) e.m.f. induced in rod AB is B v
2 3
Solution (AD):
u vcot
v 2v
sin 3
Consider a imaginary conducting loop ABC.
e.m.f. in this loop = motional e.m.f. in the rod
B d(xy)
2 dt
B dy B vcos2[yv xu] u
2 dt 2sin
5. The system shown in figure is in
equilibrium well above the ground. When the string is cut at t = 0, the
time at which the spring is at its
natural length will be (Take 2 10)
(A) 1
s6
(B) 1
s2
(C) 1
s3
(D) 2
s3
Solution (AB):
T 2k
Now,
T 3T 5Tt , , etc
4 4 4
6. Choose CORRECT statement(s) regarding X-
ray: (A) If the target material in a Coolidge tube is
changed keeping the acceleration voltage
2
constant the minimum wavelength of the
x-rays produced will change.
(B) X-ray do not get deflected on application of electric or magnetic fields
(C) X-ray can ionize a gas
(D) X-ray can be reflected, diffracted and
polarized-under suitable condition.
Solution (BCD): Conceptual
7. A hypothetical experiment conducted to
determine Young’s modulus formula x
3
.TY .
If the errors found in , ,T and Y
are 1%, 2%, 1% and 5% respectively
then the possible value(s) of x can be (A) x=0 (B) x=4
(C) x=6 (D) x=10
Solution (ABCD):
Y T 3 T
xY T
5 x 2 3 x 0,4,6,10
8. A circular conducting loop of
radius r0 and having
resistance per unit length
as shown in the figure is
placed in a magnetic field B which is constant
in space and time. The ends of the loop are
crossed and pulled in opposite directions with
a velocity v such that the loop always remains circular and the radius of the loop goes on
decreasing, then (A) Radius of the loop changes with r as r = r0
–vt/
(B) EMF induced in the loop as a function of
time is 0e 2Bv[r vt/ ]
(C) Current induced in the loop is Bv
I2
(D) Current induced in the loop is Bv
I
Solution (ABD):
If x is radius of loop, then d(2 r)
2vdt
dr v
(constant)dt
0
r t
0 0
r 0
v vt vtdx dt r r r r
Emf 2d d dr
B r 2B rdt dt dt
0vt v
2B r
2B rv Bv
current2 r
9. In the diagram shown, a
light ray is incident on
the lower medium boundary at an angle
45° with the normal.
Which of the following
statement is/are true?
(A) If 2 2 then angle of deviation is 45°
(B) if 2 2 then angle of deviation is 90°
(C) If 2 2 then angle of deviation is 135°
(D) If 2 2 then angle of deviation is 0°
Solution (AB):
For ci i , TIR will occur and then
180 2 45
For ci i ray will finally graze on upper
surface
10. A point object is kept at (1, 0, 0). A circular
plane mirror of radius 1m is kept in yz-plane such that its centre is at the origin. The
reflecting side faces positive x-axis. At which
of the following points can the image of the
object be seen? (A) ( ( 0.5, 0, 0.5) (B) (2, 2, 2)
(C) (1,1.5,1.5) (D) (1, 1,1.5)
Solution (BD):
Image cannot be seen in the region with sign
SECTION 3:
1. Match the following: COLUMN I
(A) Vd – Vc (in volt)
(B) Va – Vb (in volt)
(C) work done by 12V cell is (in J)
(D) work done by 24V cell is (in J)
COLUMN II (P) –8
(Q) +4
(R) –192
(S) 384
Solution (A-Q; B-P; C-R; D-S):
In the loop q q
24 12 04 2
3q
12 q 16 C4
1m
1m
B
vv
3 2
2
1=245°
3
(A) d cq 16
V V 4voltC 4
(B) A b A Bq q
V V V V 8volt2 2
(C) work done qE 16 12 192
(D) work done 24 16 384J
2. In a string a standing wave is set up whose
equation is given as y = 2A sin kx cos t. The
mass per unit length of the string is .
COLUMN I
(A) at t = 0
(B) at T
t8
(C) at T
t4
(D) at T
t2
COLUMN II
(P) Total energy per unit length at x=0 is 2 22 A .
(Q) Total energy per unit length at x /4 is
2 22 A .
(R) Total energy per unit at x is 2 22 A .
(S) Power transmitted through a point at
x is 0.
(T) Power transmitted through a point at
x /4 is 0.
Solution (A-PRST; B-ST; C-QST; D-PRST):
Total energy per unit length is 2 21
(2A) .2
For (A): At t=0
Power F. (at extreme, =0 for all points)
For (B): At T
t8
Power F. ( = 0 at node)
For (C): At T
t8
Power F. (F for all points)
For (D): At T
t2
Power F. (at extreme, 0 for all points)
SECTION 4:
1. A 50 kg man is standing at one end of
a 3m long plank of
mass 100 kg. The
plank is sliding towards the right with
uniform velocity of 2 m/s on the frictionless
horizontal ground. The man walks to the other end of the plank in 5s. Find distance
traveled by the man relative to the ground in
this 5 s interval. Solution (8):
11(50 100)2 50(v 3/5) 100v v
5
M/g P/g M/p M/g
M/g
11 3 8; ;
5 5 5
S 5 8m
2. A light of wavelength 3540Å falls on a metal
having work function of 2.5 eV. If ejected
electron collides with another target metal
inelastically and its total kinetic energy is utilized to raise the temperature of target
metal. The mass of target metal is 10–3 kg
and its specific heat is 160 J/kg/°C. If 1018
electrons are ejected per second, then find the
rate of raise of temperature (in °C/s) of the metal [Assume there is no loss of energy of
ejected electron by any other process, all the
electron are reaching the target metal with
max kinetic energy and take hc = 12400 ev-Å] Solution (1):
Energy of 1 photon hc 12400
eV 3.5eV3540
Thus K.E.max of e ejected = 1 eV
Thus energy lost per electron = 1 eV
= 191 1.6 10 J
Total energy gained by target metal per
second 19 181.6 10 10 J 0.16J
Thus 3
dT dT 0.160.16 ms °C/sec
dt dt 10 160
3. An electron and a proton are separated by a large distance and the electron approaches
the proton with a kinetic energy of 4.11 eV. If
the electron is captured by the proton to form
hydrogen atom in the ground state, the
wavelength of photon given off is 210 Å?
Fill the value of in your OMR sheet. Solution (7):
Since initially electron and proton are
separated by large distance potential
energy = 0
Thus when electron and proton combine to
form a hydrogen atom E (13.6 4.11)eV
hc 12400E Å 700Å
17.71
4
2m m
l
2020
W E
N
S
1045
4. An open organ pipe containing air resonates
in fundamental mode due to a tuning fork.
The measured values of length (in cm) of the
pipe and radius r (in cm) of the pipe are
94 0.1, r 5 0.05. The velocity of the
sound in air is accurately known. The
maximum percentage error in the
measurement of the frequency of that tuning
fork by this experiment is given by 2% . Find
the value of 10 .
Solution (4):
v
f2( 2e)
where e = end correction = 0.6r
v v
f2( 2 0.6r) 2( 1.2r)
f v ( 1.2r)
f v 1.2r
v 1.2 r
v 1.2r
Here v
v
= 0 (given)
f100
f
1.2 r
1001.2r
For maximum % error : = 0.1, r = 0.05
max
f 0.1 1.2 0.5100 100
f 94 1.2 5
0.16%
5. A horizontal plane with the coefficient of
friction µ supports two bodies: a bar and an electric motor with a battery on a block. A
thread attached to the bar is wound on the
shaft of the electric motor. The distance
between the bar and the electric motor is
equal to l. When the motor is switched on, the bar, whose mass is twice as great as that of
the other body, starts moving with a constant
acceleration w. The bodies collide at
t(3w µg)
nl , n here is?
Solution (2): For Bar
T 2mg 2mw
T mg mw
mg mw 2mw
w g 2w
So, 21 2(w w )t t
2 w w
2
tg 3w
6. In a car race, car A takes time t less than car
B and passes the finishing point with a
velocity v more than the velocity with which
car B passes the point. Assuming that the
cars start from rest and travel with constant
accelerations 9 m/s2 and 4 m/s2
respectively. Then the value of v/t is n. Find n Solution (6):
B Bv v 0 9(t t)
2 2
B B1 1
9(t t) 4t2 2
B B B3(t t) 2t t 3t
B Bv 0 4t
B Bv v 9(t t)
B Bv 4t 9t 9t
Bv 5t 9t
v 15t 9t 6t
v/t 6
7. Two ships A and B are 10 km apart on a line running south to north. Ship A farther north
is streaming west at 20 km/hr and ship B is
streaming north at 20 km/hr. Their distance
of closest approach is n 2 km. Find n ?
Solution (5):
Velocity of B w.r.t. A = ˆ ˆ20j 20i
Closest distance
10sin45
5 2
8. A charged particle is moving
with constant speed v in x-y
plane in a straight line as
shown in figure. Suddenly a
uniform magnetic field is switched on along y direction when particle is
at origin. The value of (in degree) so that
the particle passes through point P(0, ,
2 3 ) in the minimum possible time is .n
Find n?
Solution (3): Since x is zero and z coordinates is non zero
so particle has completed half circle, So
T m
vcos vcos2 qB
… (i)
mvsin
2 3 2R 2qB
… (ii) From (i) and (ii)
3
cos . tan 3sin 3
5
5
CHEMISTRY (PAPER-1)
SECTION 2:
1. Select the incorrect statement(s):
(A) 1IE of nitrogen atom is less than 1IE , of oxygen atom
(B) Electron gain enthalpy of oxygen is less negative than selenium
(C) Electronegativity an Pauling scale is 2.8 times the electronegativity an Mulliken scale.
(D) 6Cr is smaller then
3Cr .
Solution (AC):
2. For 3Mn
pairing energy is 28000 1cm
, o for 3
6[Mn(CN) ] is 38500
1cm then which of the
following is/are correct:
(A) Complex will be coloured (B) Complex will be low spin complex
(C) Net 1CFSE 33600cm (D) Complex will be colourless
Solution (BCD):
7cb
1259 10 em 259 nm
38500
(u.V region) and 2,1,1 0,02gt , eg
3. Select the correct statement with respect to the p d dative bond:
(A) In 3 2(Ph Si) O, the Si O Si group is nearly linear.
(B) Silanols such as ( 3 3(CH ) SiOH are stronger protonic acids then their carbon analogs
(C) Trisilysl phosphine 3 3(H Si) P is planar
(D) 4ClO does not polymerise at all
Solution (ABD):
(A) steric repulsions of bulkier groups and p d dative bonding favour for linear Si O Si
groups
(B) Due to stabilization of the conjugate base only by O(p ) Si(d ) bonding motion.
(C) It is pyramidal because p d bonding is not effective due to bigger size of phosphorous
atom
(D) Mont effective, p d overlapping due to small size of chlorine
4. The reaction 22NO Br 2NOBr follows the mechanism;
1. fast
2 2No Br NOBr
2. slow
2NOBr NO 2NOBr
Which of the following is/are true regarding this:
(A) The order of reaction with respect to NO is two (B) The molecularity of the steps (1) and (2) are two each.
(C) The moleuclarity of the overall reaction is three.
(D) The overall order of reaction is three.
Solution (ABD):
5. 10 moles of a liquid are 50% converted into its vapour at its boiling point o(273 C) and of a pressure of
1 atm. If the value of latent heat of vaporization of liquid is 273 L atom/mole then which of the
following statement is/are. Correct: Assume volume of liquid to be negligible and vapour of the liquid to
behave ideally:
(A) work done by the system in the above process is 224 L atm
(B) the enthalpy change ( H) for the above process is 1365 L atm (magnitude only)
(C) the entropy of the system increases by 2.5 L atm in the above process
(D) the value of U for the above process is 1589 L atm
Solution (ABC):
fV 5 R 546 224L
ext.W P ( V) 1atm(224L) 224 L atm
Work done by system 224L atm
6
OH
OH
O
OHOH
OOO
COOHOH
HOOC
OH
2CH I
OHOOC
O
Enthalpy change ( H) Q 273 5 1365 L atm
vap.H 1365S 2.5L atm/k
T 546
U q w 1365 224 1141L atm
6. Select the correct statement(s) related to hexagonal close packing of identical sphere in three
dimensions: (A) In one unit cell there are 12 octahedral voids and all are completely inside the unit cell
(B) In one unit cell there are six octahedral voids and all are completely inside the unit cell
(C) In one unit cell there are six octahedral voids and out of which three are completely inside the unit
cell and other three are from contribution of octahedral voids which are partially inside the unit cell
(D) Co-ordination number of every sphere is 12 in hcp lattice Solution (BD):
hcp AB AB AB.......... pattern repeat
For calculating voids between two layers A and B octahedral void=3
So total octahedral voids=6=all are completely inside.
7. Which of the following statements are correct:
(A) Average velocity of molecules of a gas in a container is zero
(B) All molecules in gas are moving with the same speed
(C) If an open container is heated from 300 K to 400 K the fraction of air which goes out with
respect to originally present is 1/4
(D) It compressibility factor of gas at STP is less than unity then its molar volume is less than 22.4l at STP
Solution (ACD):
C- 1 1 2 2n T n T Escaped out
3n n
14n 4
2n 300 n 400
23
n n4
D- PV
1 Z 1RT
Molar volume for Z 1 is 22.4 littre so for Z 1 molar volume is less than 22.4 litre at STP.
8. The products of the following reaction can be:
(2) CH I(i) CHCl /NaOH (3) (i) AgNO NH OH2 23 3 4
(ii) HP
(A) (B) (C) (D)
Solution (AC):
OHOH
3(1) CHCl /NaOH
Re imer Tiemann reaction
OHC
ONa
ONa
+
CHO
ONa
ONa
CHO
+2 2CH I
O
O
O
O
CHO(3)
COOH
+
O
O
O
O
COOH
7
7
2H C14
2C H
O
182H /H O
9.
Which cannot be the product:
(A)
14
2 218
H C CH| |
HO HO
(B)
14
2 2
18
CH CH| |
HO OH
(C)
2 218 18
CH CH| |OH OH
(D)
14
2 2CH CH| |OH OH
Solution (CD):
Cis-(1R-3S)-disec-butylcyclobutane
10. What is/are true about the following compound:
(A) it has centre of symmetry (inversion centre) (B) It has plane of symmetry
(C) it does not have two fold axis 2(C ) of symmetry (D) It is an chiral molecule
Solution (BCD):
It is a meso achiral compound and it has no. centre of symmetry and no. axis of symmetry but plane of
symmetry.
SECTION 3:
1. Match the following column I & column II.
Given o
2Cu /CuE 0.34V,
o
Cl /Cl2
E 1.36V,
o
Br /Br2
E 1.08V ,
o
I /I2
E 0.54V
COLUMN I COLUMN II
(A) 2
2Cu 2Cl Cu Cl (P) Can produce electricity in the galvanic cell
(B) 2
2Cl Cu Cu 2Cl (Q) Can be made to occur in electrolytic cell
(C) 2I starch solution + chlorine water (R) appearance of brown colour
(D) 42Br CCl chlorine water (S) Appearance of violet colour
Solution (A-Q, B-P, C-S, D-R): o ored red 2Cl /Cl Cu /Cu2
(E ) (E )
A Q So reduction of 2Cl to Cl will occur at cathode and oxidation of Cu to 2Cu
will occur at
anode. So reverse reaction is possible it means such revere reaction can be made to occur in an
electrolysis cell.
B P as mentioned in (a) the reaction is feasible and hence can produce electricity in the galvanic
cell.
C S 2 22I Cl I (violet) 2Cl
182H O14
2 2H C CH
O
H14 14
2 2 2 2CH CH CH CH
18HO OH HO 18OH
+
a
b
3H C3CHH
5 2H C
Cis-(1R-3S)-disec-butylcyclobutane
H
2 5C H
8
NH NH+
2CH
2NO 2CH
O
+
o ored redCl /Cl Br /Br2 2
(E ) (E )
D R 2 22Br Cl Br (brown) 2Cl
o ored redCl /Cl Br /Br2 2
(E ) (E )
2. Match the following column I & column II:
COLUMN I COLUMN II
(A) (P) Stabilised by hyperconjugation
(B) (Q) Stabilised by -I effect
(C) (R) Stabilised by +M effect
(D) (S) Stabilised by M effect
Solution (A-PR; B-QS; C-P; D-R)
SECTION 4:
1. Electrons in a sample of H-atoms make. Transitions from state n=x to some lower excited state. The emission spectrum from the sample is found to contain only the lines belonging to a particular series. If
one of the photons had on energy of 0.6375eV. Then find the value of x 3
(take 0.6375eV 0.85eV)4
Solution (8):
We have 3
E 0.85eV4
as energy 0.6375 the photon will belong to Brackett series (as for
Brackett 0.31 E 0.85)
2 2
1 1 10.85 1 13.6
4 4 n
21 13.6 4
0.85 1 14 16 n
n 8
2. At 2 atm. pressure the value of x/m will be where x is the mass gas adsorbed per unit mass of
adsorbent according to given graph:
Solution (4):
x 1log logK logP
m n
|aK |0/2 K 2 nx
K(P)m
11
n n 1
xx2 2 4
m
log 2
log P
o45log x/m
9
9
O
O
H O
Br Br
Br
2 5OC H
3. 3NH is heated initially of 15 atm from o27 C to
o127 C of constant volume. At o127 C equilibrium is
established. The new pressure at equilibrium at o127 C becomes 30 atm for the reaction
3 2 22NH (g) N (g) 3H (g). Then find the fraction of moles of 3NH actually decomposed.
(represent an … 1x 10 )
Solution (5):
3 2 2a 0 0
a 2x x 3x
2NH N 3H
2 212
1 2
P PP 15P 20 atm
T T 300 400
Now; a 2x 30 1
x aa 20 4
Fraction of 32x 1
NH 0.5a 2
4. Among the following complexes how many have ‘spin only’ magnetic moment d 2.83 BM? 2 2 2 4
4 4 2 3 2 4 6 6[Ni(CO) ], [Ni(CN) ] , [NiCl (PPh ) ], [NiCl ] ,[NiF ] , [NiF ] ,
2 2 23 6 3 2 6[Ni(NH ) ] ,[Ni(en) ] ,[Ni(H O) ]
Solution (6): 2 4 2 2 2
2 3 2 4 6 3 6 3 2 6[NiCl (PPh ) ],[NiCl ] ,[NiF ] ,[Ni(NH ) ] [Ni(en) ] ,[Ni(H O) ]
5. (i) CH MgBr(3equivalent)3
(ii) H /H O2
product
How many bromine atoms are present in final major product?
Solution (2):
6. Glucose is subjected to bond cleavage by 4HIO . The number of formic acid unit(s) formed per unit of
glucose is…..
Solution (5):
O
O
H O
Br Br
2 5C OC H(i) CH MgBr(3equivalent)3
(ii) H /H O2
OH
Br
2 5COOC H
CH3CH
Br Br OH
2CH OH
HO
H
H
OH
OH
H
O
H OH
O O
C H
5H C OH 1H C H
10
3CH3CH
OH OH
Ph Ph
7. In the stander dization of 2 2 3Na S O using 2 2 7K Cr O by iodometry the equivalent weight of 2 2 7K Cr O is
molecular weight
x
Solution (6):
2 32 27
2
2 32 2 27
Cr O 14H 6e 2Cr 7H O
(2I I 2e ) 3
Cr O 14H 6I 2Cr 3I 7H O
So eq. weightmoleuclar weight
6
8. Conc. H SO .2 4 Products
How many number of different type of carbonyl products (only structural isomers) can be formed (major
or minor) in this reaction, (considering all type of possible migrations): Solution (4):
Pincol pinacolone rearrangement.
11
MATHEMATICS (PAPER 1) SECTION 2:
1. If the roots of the equation
2 2c x c a b x ab 0 are sinA, sinB where
A, B and C are the angles and a, b, c are the
opposite sides of the triangle ABC, then
(A) the triangle is right angled
(B) the triangle is acute angled
(C) the triangle is obtuse angled
(D) sinA cos A a b /c
Solution (AD):
2
c a b sinA sinBsinA sinB
sinCc
sinC 1 oC 90 oA B 90
cos A sinB
Hence sinA cosA sinA sinB a b
c
.
2. The vectors a,b & c are of the same length
and taken pair-wise, they form equal angles.
If ˆ ˆa i j and ˆ ˆb j k the coordinates of c
are
(A) 1,0,1 (B) 1,2,3
(C) 1,1,2 (D) 1 4 1
, ,3 3 3
Solution (AD):
ˆ ˆa i j and ˆ ˆb j k a b 2 .
Hence c 2 .
Let 1 2 3ˆ ˆ ˆc c i c j c k .
2 2 21 2 3c c c 2 2 2 2
1 2 3c c c 2
Also ˆ ˆ ˆ ˆ ˆ ˆa.b i j 0k . 0i j k
a b cos 0 1 0 2. 2 cos 1
1
cos2
1 2 3ˆ ˆ ˆ ˆ ˆa.c i j . c i c j c k
1 2 3ˆ ˆ ˆ ˆ ˆa . c cos i j . c i c j c k
1 21
2. 2. c c2 1 2c c 1
Similarly 2 3c c 1
1 2 3 2c 1 c ,c 1 c
2 2 21 2 3c c c 2
2 222 2 21 c c 1 c 2
22 23c 4c 2 2 2
4c 0 or
3
For 2c 0 : 1 3c 1, c 1 ˆ ˆc i k
c 1,0,1
For 24
c3
: 1 31 1
c , c3 3
1 4 1ˆ ˆ ˆc i j k3 3 3
1 4 1
, ,3 3 3
3. If 1 2 nA ,A ,....,A are n independent events
such that i1
P A ,i 1,2,...,ni 1
. The
probability that none of 1 2 nA ,A ,...,A occurs
is
(A) n
n 1 (B)
1
n 1
(C) less than1
n (D) greater than
1
n
Solution (BC):
Since 1 2 nA ,A ,...,A are independent events
therefore
/ / /2 nP A A ... A / / /
n1 2P A P A ....P A
1 1 1
1 1 ... 12 3 n 1
1 2 3 n 1 n 1 1
... .2 3 4 n n 1 n 1 n
4. Number of integral values of ‘b’ for which
inequality 2 2a 1 x 4 a b x 2 0 is true
for atleast one ‘x’ & a R , is
(A) 0 (B) 1
(C) 2 (D) 3
Solution (A):
Since 2 2a 1 x 4 a b x 2 0 , for at
least one x, hence discriminant must be
positive.
2 24 a b 4 a 1 2 0 a R
2 2 216 a b 2ab 8 a 1 0 a R
2 2 22 a b 2ab a 1 0 a R
2 2a 4ab 2b 1 0 a R
2 2a a 4b 2b 1 0 a R
Since above quadratic in ‘a’ must be positive a R , hence its discriminant must be
negative.
2 24b 4 1 2b 1 0
2 216b 8b 4 0 28b 4 0 , which is
not possible. Hence number of possible
integral values of b is 0.
5. Consider a square on the complex plane. The
complex numbers corresponding to its four vertices are the four distinct roots of the
equation with integer coefficients
12
4 3 2x px qx rx s 0 , then the minimum
area of the square is
(A) 1
2 (B) 1
(C) 2 (D) 2
Solution (C):
Let origin be the centre of the square and edge
length be 2a units.
Area 22a 2a 4a
As per question:
a ai, a ai, a ai & ia a are root of the
equation: 4 3 2x px qx rx s 0
Product of the roots 4 s
P 11
a ai a ai a ai a ai
s a 1 i a 1 i a 1 i a 1 i
4s a 1 i 1 i 1 i 1 i
4s a 2 2 4s 4a
Since s is an integer, therefore minimum
value of 4a for which 44a is an integer is
4 1a
4 .
Area 2 4 14a 4 a 4
4
14 2
2 sq.units.
6. If is the imaginary cube root of unity such
that
n r n r
P 1 P 1
r 1 p 1 r 1 p 1
r. 155 r. 155
, where n is multiple of 3, then possible
values of n can be
(A) 27 (B) 30
(C) 33 (D) 36 Solution (B):
n r
p 1
r 1 p 1
r. 155
n
o 1 2 r 1
r 1
r .... 155
1 0 1 0 1 21 2 3
0 1 2 3 0 1 2 n 14 ... n ...
155
n1 4 7 .... terms
3
2 2n2 5 8 ... terms 155 1
3
2n n n2 1 1 3
3 2 3 3 2
2n2 2 1 3 155 155
3
2 2n n2 n 3 4 n 3 155 155
6 6
2 2n nn 1 n 1 155 155
6 6
2n nn 1 155 155 n 1
6 6
As per question:
2n nn 1 155 155 n 1
6 6
2n nn 1 155 155 n 1
6 6
2n nn 1 155 155 n 1
6 6
is purely real
n
155 n 1 06
155 6 n n 1
2n n 155 6 0 n 30 n 31 0
n 30, 31
From the given options only option (B) is
correct.
7. If the grid shown below consists of n rows
then number of triangles orienting the same way as ABC are
(A) n 12C
(B) n 23C
(C) n 13C
(D) n 22C
Solution (C):
There is one triangle of height n, and 1 2
triangles of heights n 1 (these are AQR,
PBN, LMC). Similarly there are 1 2 3
triangles of height n 2 , and 1 2 ...n
triangles of height 1. Hence the number of
triangles in the grid the same way as ABC are
1 1 2 1 2 3 ... 1 2 ... n .
n 1 n 1 2 n 2 3 ... 1 n
n
r 1
n r 1 r
13
n n n n
2 2
r 1 r 1 r 1 r 1
n 1 r r n 1 r r
n n 1 n n 1 2n 1
n 12 6
n n 1 2n 1n 1
2 3
n n 1 3n 3 2n 1
2 3
n n 1 n n 1 n 2n 2
2 3 6
n 23C
8.
n
nn
3n 1lim
4n 1
is equal to
(A) 3
4 if n is even (B) 0 if n is even
(C) 3
4 if n is odd (D) none of these
Solution (AC):
CASE I: If n is even, say n 2k. Then the
limit is:
2k
2kk k
6k 1 6k 1lim lim
8k 18k 1
k
16
6 3klim1 8 4
8k
CASE II: If n is odd, say n 2k 1. Then the
limit is:
2k 1
2k 1k k
3 2k 1 1 6k 3 1lim lim
8k 4 14 2k 1 1
k
46
3klim5 4
8k
n
nk
3n 1 3lim
44n 1 1
, n even or odd.
9. Consider a circle with its centre lying on the
focus of the parabola 2y 2px such that it
touches the directrix of the parabola. Then
point(s) of intersection of the circle and the
parabola is/are:
(A) p
,p2
(B) p
, p2
(C)p
,p2
(D)p
, p2
Solution (AB):
2y 2px 2 py 4 x
2
…(i)
Hence focus is at p
,02
& equation of
directrix is p
x2
2x p 0 . Since the
circle touches the directrix therefore the
radius r of circle is:
2 2
p2 p
2p2r p
22 0
Hence equation of the circle is:
22 2p
x y p2
2
2 2 3px y px 0
4
…(ii)
Eliminating y from (i) and (ii):
2
2 3px 2px px 0
4 2 23
x px p 04
2 2p p 3p p 2px
2 2
p 3p
x ,2 2
p
x2
, (x cannot be negative)
2 py 2p
2
y p .
Hence the points of intersection are
p p,p , , p
2 2
.
10. A variable point P on the ellipse 2 2
2 2
x y1
a b is
joined to its foci S & /S . Let e be the
eccentricity of this ellipse. The correct
statements are (A) The locus of the incentre of the triangle
/PSS is the ellipse
2 22
2 2 2 2
e 1 yx1
a e b e
.
(B) The locus of the incentre of the triangle /PSS is the hyperbola
2 22
2 2 2 2
e 1 yx1
a e b e
.
(C) The eccentricity of the locus of the
incentre of the triangle /PSS is
1/21 e
2e
.
(D) The eccentricity of the locus of the
incentre of the triangle /PSS is
1/22e
1 e
.
14
Solution (AD):
Let the point P be acos ,bsin .
Also S ae,0 and /S ae,0
PS a ex a aecos /PS a ex a aecos /SS 2ae
Hence coordinates h,k of incenter of the
triangle /PSS are given by:
1 2 3ax bx cxh
a b c
2ae a cos a aecos ae
a aecos ae
2ae a aecos a aecos
2 22ae a cos 2a e cos
2ae 2a
2aecos ae cos
e 1
aecos
1 2 3ay by cyk
a b c
2ae bsin a aecos 0
a aecos 0
2ae a aecos a aecos
2ae bsin besin
2ae 2a e 1
k e 1h
cos & sinae be
Eliminating :
22h e 1
k 1ae be
222
2 2 2 2
y e 1x1
a e b e
.
Hence option (A) is correct. Also eccentricity /e of the above ellipse is given by:
2 2
2/
2 2
b e
e 1e 1
a e
2
22
b1
a e 1
2
2
1 e1
e 1
,
22
2
b1 e
a
2
2
2e 2e 2e
e 1e 1
Hence option (D) is correct.
SECTION 3:
1. Match the following:
‘n’ whole numbers are randomly chosen and multiplied.
COLUMN-I COLUMN- II
(A) The probability that the (P)n n
n
8 4
10
last digit is 1, 3, 7 or 9 is
(B) The probability that the (Q)n n
n
5 4
10
last digit is 2, 4, 6, 8 is
(C) The probability that the (R)
n4
10
last digit is 5 is (D) The probability that the
last digit is zero is
(S) n n n n
n
10 8 5 4
10
Solution (A R, B P, CQ, D S):
(A) The required event will occur if last digit
in all chosen numbers is 1, 5, 7 or 9.
Required probability =
n4
10
(B) The required probability is equal to the
probability that last digit is 2, 4, 6, 8
n n
n
8 4P 1,2,3,4,5,6,7,8,9 P 1,3,7,9
10
(C) n n
n
5 4P 1,3,5,7,9 P 1,3,7,9
10
(D) n n n nP 0,5 P 5 10 8 5 4
n n n n
n
10 8 5 4
10
2. Match the following:
COLUMN I COLUMN II
(A) If x 1 & 1 2sin cot x 1 dx
(P) 0
f x k , where f 1 0 , then
the value of 2f e is equal to
(B) If
2
2012
sec xdx g x C
tan x tanx
, (Q) 1
where n2
g4 2011
, then
x
2
lim g x
is equal to
(C) If the function x
f xx
(R) 2
is self inverse, then
can be
15
(D) Given x 0,1 & (S) even number
x
20
dtf x
1 t
,
then the value of
22f x f 1 x
is equal to
Solution (ARS, B PS, C PQR, DQ):
(A) 1 2sin cot x 1 dx f x k
1
dx f x kx
lnx C f x k
f x lnx & k C
2 2f 1 0 & f e lne 2
(B)
2
2012
sec xdx g x C
tan x tanx
Let tanx t 2sec x dx dt
2012
dtg x C
t t
2012
2011
t dtg x C
1 t
Let 20111 t y
20122011 t dt dy
2012 1t dt dy
2011
dy
g x C2011 y
1
ln y C g x C2011
20111ln 1 t C g x C
2011
20111ln 1 tan x C g x C
2011
20111g x ln 1 tan x
2011
2011
2011
1 cos xln 1
2011 sin x
x
2
1 0lim g x ln 1
2011 1
1 1
ln1 0 02011 2011
.
(C) x
f xx
Let x
yx
xy y x xy x y
y
xy 1
1 xf x
x 1
Since f x is self in verse.
x x
, x Rx x 1
2 2 2x x x x x x
x R
2 2x 1 x 1 1 0
x R
21 0 & 1 0 & 1 0
1 & R
(D) x
20
dtf x
1 t
x
1
0
f x sin t
1f x sin x
2 1 2f 1 x sin 1 x
2 1f 1 x cos x
2 1 12 2f x f 1 x sin x cos x
2
12
.
SECTION 4:
1. Given that the vectors a and b are non-
collinear. The value of x y for which the
vector equality 2u v w holds, where
u xa 2yb, v 2ya 3xb, w 4a 2b , is
Solution (2):
2u v w
2 xa 2yb 2ya 3xb 4a 2b
2x 2y 4 a 3x 4y 2 b 0
2x 2y 4 0 and 3x 4y 2 0
a,b are linearly independent
10 4
x ,y7 7
x y 2 .
2. Let N be the number of natural numbers
which are smaller than 82 10 and which can
be written by means of the digits 1 and 2. The
value of N 760 is
Solution (6): 82 10 200000000, which is nine digit
number. Hence all the numbers of 1 digit, 2
16
digit, …, 8 digits formed by 1 and 2 are clearly
smaller that 82 10 . Also all the 9 digit
numbers beginning with 1 are smaller than 82 10 .
Number of one digit numbers 12 2
Number of two digit numbers 22 2 2
Number of three digit numbers 32 2 2 2
Number of eight digit numbers 82
Number of nine digit numbers beginning with one (the first place is to be filled with 1 & remaining 8 places can be filled in 2ways viz:
1 & 2) 8 81 2 2 .
Hence total number of required numbers
8
2 3 8 8 82 2 1
N 2 2 2 ... 2 2 22 1
9 82 2 2 512 2 256 766 .
N 760 6 .
3. The value of 11 12sin 11 sin 126 6
19 20... sin 19 sin 206 6
is
Solution (0): 11 13 15 17 1911 ,13 ,15 ,17 ,19 are odd and
12 1412 ,14 , 16 18 2016 ,18 ,20 are even. Hence
For n 11,13,15,17,19
n 1sin n sin
6 6 2
& for
n 12,14,16,18,20. n 1sin n sin
6 6 2
Hence the required sum
1 1 1 1 1 1 1 1 1 10
2 2 2 2 2 2 2 2 2 2 .
4. The value of k for which the function
4x
2 2
2
e 1f x
x xsin log 1
2k
, x 0 , f 0 8
may be a continuous function at x 0 is
(given that k 0 ):
Solution (2):
4
x
2 2x 0
2
e 1lim
x xsin log 1
2k
2 24
x 22
2 2x 0
2
x x
e 1 2klim 2kx x x
sin log 12k
4 2 2 21 1 1 2k 2k
For continuity at x 0 :
x 0
f 0 lim f x
28 2k k 2
5. If the product of the perpendiculars drawn
from the point 1,2 to the pair of lines
2 2x 4xy y 0 is P then 13
4P is equal to
Solution (1):
We know that the product of perpendiculars
drawn from 1 1x ,y to the lines represented by
2 2ax 2hxy by 0 is
2 21 1 1 1
2 2
ax 2hx y by
a b 4h
.
Here 1a 1,2h 4,b 1,x 1 , 1y 2 .
Required product
2 2
2 2
1 1 4 1 2 1 2 13P
41 1 4 2
13
14P
6. If g x
30
dtf x
1 t
, where
cos x
2
0
g x 1 sint dt then the value of
/1 f2
is equal to
Solution (0):
g x
30
dtf x
1 t
//
3
g xf x
1 g x
…(i)
Also cos x
2
0
g x 1 sint dt
/ 2g x 1 sin cos x sinx …(ii)
Hence from (i) and (ii):
2
/
2
1 sin cos x sinxf x
1 g x
…(iii)
Also cos
22
0
g 1 sint dt2
0
2
0
1 sint dt 0
…(iv)
17
2
/
2
1 sin cos sin2 2
f2
1 g2
2
1 sin01
1 0
/1 f 02
7. The number of zeros in the end in the product
of 6 7 8 9 515 6 7 8 ... 50 must be
Solution (356):
Since, one pair of 2 and 5 makes 10 (i.e.,
2 5 10 ) which gives one zero at the end.
Here,
6 11 16 21 26 315 ,10 ,15 ,20 ,25 ,30 ,
41 46 5140 ,45 ,50
Total numbers of fives
6 11 16 21 26 31 36 41 46 51 26 51
10
6 51 772
356
Since, number of s2 is greater than that of 5.
Thus, number of zeros at the end 356 .
8. In a ABC , the maximum value of
2 Aa cos
21000
a b c
must be
Solution (750):
2 2 2A B Ca cos bcos ccos
2 2 2
a b c
a 1 cos A b 1 cosB c 1 cosC
2 a b c
a b c a cos A bcosB ccosC
2 a b c
1 a cos A bcosB ccosC
2 4s
1 R
sin2A sin2B sin2C2 4s
a b c2R
sinA sinB sinC
1 R
4sinAsinBsinC2 4s
2
1 R a b c 1 abc4. . .
2 4s 2R 2R 2R 2 9R s
2
1 4R 1 r 1 r 1 11 1
2 2 2R 2 R 2 28R s
r 1
R 2
2 2 2A B Ca cos bcos ccos
32 2 2
a b c 4
2 2 2A B Ca cos bcos ccos
2 2 21000
a b c
31000 750
4
**************
18
IO
Principle
axis
IO
Principle
axis
PHYSICS (PAPER-2) SECTION 1:
PARAGRAPH FOR NEXT TWO QUESTIONS:
A light of wavelength is incident on a metal
sheet of work function 2eV. The wavelength
varies with time as = 3000 + 40t, where is
in Å and t is in second. The power incident on metal sheet is constant at 100 W. This signal is
switched on and off for time intervals of 2
minutes and 1 minutes respectively. Each time
the signal is switched on, the start from initial
value of 3000Å. The metal plate is grounded and
electron clouding is negligible. The efficiency of
photoemission is 1% (hc = 12400 eVÅ, 1eV = 1.6
× 10–19 J) 1. The time after which photoemission will stop
is
(A) 120 s (B) 80 s
(C) 60 s (D) 180 s Solution (B):
wf 00
hc6200 Å
for emission
0 3000 40t 6200 t 80sec
2. The variation of rate of emission vs time is
(A) (B)
(C) (D)
Solution (B):
dN P 100
(3000 40t)dt hc hc
PARAGRAPH FOR NEXT TWO QUESTIONS In the shown circuit the key is shifted from position 1 to position 2
3. Final charge on capacitor B will be
(A) 2 coulomb (B) 4 coulomb (C) 1 coulomb (D) zero
Solution (A): Charge on B will not charge due to shifting of
key because the circuit becomes open
4. Extra work done by 2 volt cell due to shifting
of key is
(A) 8 J (B) zero
(C) 12 J (D) 4 J
Solution (C): Charge through cell = 6 coulomb
Work done = qE = 12 joule
SECTION-2:
1. The figure shows
positions of object O
and its diminished
image I. This is
possible if (A) a concave mirror is placed to the right of I
(B) a convex mirror is placed between O and I
(C) a concave lens is placed to the right of I
(D) a concave lens is placed between O and I
Solution (BC): Conceptual
2. Two pulses on the same string are described
by the following wave equations:
1 22 2
5 5y and y
(3x 4t) 2 (3x 4t 6) 2
where x and t are in m and second
respectively. Choose the correct statement.
(A) Pulse y1 and pulse y2 travel along positive
and negative x-axis respectively. (B) At t=3/4s, displacement at all points on
the string is zero.
(C) At x=1m displacement is zero for all time
(D) Energy of string is zero at t = 3/4s
Solution (ABC):
For (A): 1 1 2 2y f (x vt) and y f (x vt)
For (B): At 1 2
3t s, y y y
4
2 2
5 50
(3x 3) 2 (3x 3) 2
For (C): At x = 1m
2 2
5 5y 0
(3 4t) 2 (4t 3) 2
For (D): At 3
t s4
potential energy of string is
minimum and kinetic energy is maximum.
3. Two identical incandescent light bulbs are connected as shown in the figure. When an
AC voltage source of frequency f is applied in
circuit, which of the following observations
will be wrong
2V
A
2F 2F
B
1
2
4V
22
2V2
initial
24
2V
final
19
O I
(A) Maximum brightness
of both does not
depend on frequency
f, if 1
f 1/LC2
(B) Both bulbs will glow
with same
brightness provided frequency
1
f 1/LC2
(C) Bulb b1 will light up initially and goes off, bulb b2 will be ON constantly
(D) Bulb b1 will blink and bulb b2 will be ON
constantly
Solution (ACD): At resonance both bulbs will glow with same
brightness. At resonance,
L CX X or 1
2 fL2 fC
or 1
f2 LC
4. The diagram shows a
convex lens and an object
O placed at a distance
greater than 2f on its
optical axis. Which of the
following statements is/are true?
(A) if O is moved towards lens I will move
away from lens
(B) if O is moved away from lens I will move
towards lens (C) If lens is rotated clockwise about its pole, I
will move downwards
(D) If lens is rotated clockwise about its pole, I
will move upwards
Solution (A): In convex lens image moves in same direction as that of object. Even if less will rotate, image
will move along principle axis
5. Two concentric metallic shells of radii R and
2R are placed in vacuum. The inner shell is having charge Q and outer shell is uncharged.
If they are connected with a conducting wire.
Then
(A) Q amount of charge will flow from inner to
outer shell.
(B) Q/e number of electrons will flow from outer to inner shell, where ‘e’ is charge on
an electron.
(C)
2KQ
4Ramount of heat is produced in the
wire.
(D)
2KQ
2Ramount of heat is produced in the
wire.
Solution (ABC):
(After connection entire charge moves towards
outer shell) Q
Q ne ne
Amount of heat produced = loss in electric
potential energy 2 2 2KQ KQ KQ
2R 4R 4R
6. A hollow cylinder, a spherical shell, a solid
cylinder and a solid sphere are allowed to roll
on an inclined rough surface of coefficient of
friction and inclination . The correct
statement(s) is/ are
(A) Frictional force will be equal for all the rolling objects, if having same mass.
(B) If all the objects are rolling and have same
mass, the K.E. of all the objects will be
same at the bottom of inclined plane.
(C) Work done by the frictional force will be
zero, if objects are rolling. (D) If cylindrical shell can roll on inclined
plane, all other objects will also roll.
Solution (BCD):
mgsin f ma … (i)
2fR I mK … (ii)
K is radius of gyration
From (i) and (ii), 2 2 2
2
gRsin gsin;
(R K ) K1
R
7. A standing wave of time period T is set up in a string clamped between two rigid supports. At
t = 0 antinode is at its maximum is placement
2A.
(A) The energy density of a node is equal to
energy density of an antinode for the first time at t = T/4.
(B) The energy density of node and antinode
becomes equal after T/2 second.
(C) The displacement of the particle at
antinode at T
t is 2 A8
(D) The displacement of the particle at node is zero
Solution (CD):
20
60°
60°
30°30
°
3m
3m
3O
A
B
A
B
C
Equation of SHM of particle which is at
antinode is 2
y 2Asin tT
at time T
t8
y 2Asin 2A4
Displacement of particle at node is always
zero.
8. In a photoelectric effect experiment, if f is the
frequency of radiations incident on the metal surface and I is the intensity of the incident
radiations, then choose the correct
statement(s).
(A) If f is increased keeping I and work
function constant then maximum kinetic
energy of photoelectron increases. (B) If distance between cathode and anode is
increased stopping potential increases
(C) If I is increased keeping f and work
function constant then stopping potential
remains same and saturation current increases.
(D) If work function is decreased keeping f
and I constant then stopping potential
increases
Solution (ACD):
max wfKE hf
Saturation current intensity (I):
maxS
KEV
e
SECTION-4: 1. A non conducting ring
(of mass m, radius r,
having charge Q) is
placed on a rough
horizontal surface (in
a region with transverse magnetic field). The field is increasing with time at the rate R and
coefficient of friction between the surface and
the ring is . For ring to remain in
equilibrium should be greater than QrR
xmgx
is …
Solution (2): For ring to be in equilibrium
2r R
mgr Q .r2 r
2. Figure shows a
square loop 10
cm on each side
in the x-y plane with its centre at
the origin. An
infinite wire is at z = 12 cm above y-axis. Then
magnetic flux through ABCD loop due to
straight infinite wire is x Weber. Then value of
x is
Solution (0): Net flux through rectangular plane is zero.
3. Figure shows circular
region of radius
R 3m in which
upper half has
uniform magnetic field
ˆB 0.2 k T and
lower half has uniform
magnetic field ˆB 0.2kT . A very thin parallel
beam of point charges each having mass
m 2gm, speed v 0.3m/sec and charge
q 1mC are projected along the diameter as
shown in figure. A screen is placed
perpendicular to inclined velocity of charges
as shown. If the distance between the points on screen where positive and negative charges
will strikes is 4X meters, then calculate X.
Solution (3): When positive charge enters
mv
RqB
= 3
3
AB 3 sin602
B C BB tan60
(OA OA)tan60
3
(2 3 ) 32
3
32
4. Full scale
deflection current
for galvanometer is
1Amp. What
should be the
value of shunt resistance in so that
galvanometer shows half scale deflection.
Solution (2):
For half deflection 1
i Amp.2
s s1 38
(20 18) 9.5 R R 22 2 9.5
5. An electron is shot into one end of a solenoid.
As it enters the uniform magnetic field within
the solenoid, its speed is 800m/s and it
velocity vector makes an angle of o30 with the
central axis of the solenoid. The solenoid carries 4.0 A current and has 8000 turns
along its length. Find number of revolutions
made by the electron within the solenoid by
++++
++++++++++++++++
++++ B
Top view
21
the time it emerges from the solenoid’s
opposite end. (Use charge to mass ratio e
m
for electron 113 10 C/kg ). Fill your
answer of y if total number of revolution is
6y10 .
5
Solution (8):
70
3000B ni 4 10 4
t800 cos30
No. of revolutions 7t e 4 10 32000
2 m 800cos30 2 m
qB
11 1013 10 10 32
3800
2
3208
400
6. A potential difference of 500V is applied
across a parallel plate capacitor. The
separation between the plates is 2×10–3m.
The plates of the capacitor are vertical. An electron is projected vertically upwards
between the plates with a velocity of 105 m/s
and it moves un-deflected between the plates.
If the magnetic field acting perpendicular to
the electric field has magnitude of
2n 5Wb/m
10
, then n is: (neglect gravity)
Solution (5): qE qvB
5
3
500 50010 B B 2.5 n 5
2002 10
7. 90% of a radioactive sample is left undecayed
after time t has elapsed. The percentage of the initial sample will decay in a total time 2t is
n1.9 10 .the value of n is
Solution (1): In time ‘t’ sec 10% has decayed so in next t
sec 10% of removing i.e. 90%. i.e. 9% more %
will decay so total decay percentage is 19%
8. The K, L and M energy levels of platinum lie
roughly at 78, 12 and 3 keV respectively. If
the ratio of wavelength of K line to that of
K line in X ray spectrum is
2n
22 find n.
Solution (5):
211 2
2 1
E 25E 66,E 75,
E 22
22
22
CHEMISTRY (PAPER-2)
SECTION 1:
PARAGRAPH FOR NEXT TWO QUESTIONS:
The dissolution of ammonia gas in water does not obey Henry’s law. On dissolving, a major portion of
ammonia molecules unite with 2H O to form 4NH OH molecules. 4NH OH again dissociates into
4NH and OH ions. In solution therefore, we have 3NH molecules, 4NH OH molecules and 4NH
ions
and the following equilibrium exist:
3NH (g) (pressure P and concentration c) Initially 3 2 4 4NH ( ) H O NH OH NH OH l
Let 1 3c mol/L of NH pass in liquid state which on dissolution in water forms 2c mol/L of 4NH OH . The
solution contains 3c mol/L of 4NH ions:
1. Total concentration of ammonia, which can be determined by volumetric analysis is equal to:
(A) 1 2c c (B) 1c (C) 1 3c c (D) 2 3c c
Solution (B):
The intermediate solution of acid will react with all the 3NH present in solution
2. Degree of dissociation of ammonium hydroxide is:
(A) 1c (B) 3 1c /c (C) 3c /c (D) 3 2c /c
Solution (D):
34
2
cMole of NH OH dissociated
Total mole c
PARAGRAPH FOR NEXT TWO QUESTIONS:
3B PBr C
C NaCN D
2 4 6 10 2D H SO E(C H O )
2E SOCl F
3F AlCl G
H SOLiAlH 2 44G H I
3. Find out structure of I:
(A) (B) (C) (D)
4. The compounds D and F must be:
(A) (B)
(C) (D)
Solution (Q.3-C, Q.4-A):
23
23
N
M
Time
[A]0/2
[A]0
H SOPBr NaCN 2 43
2 2 2 2(C) (D)
Ph CH CH Br Ph CH CH CN
SECTION 2:
1. If n nr ,U and nE are radius of orbit speed of electron and energy of electron in H-like atom shell then,
which of the following are proportional to number of shell ‘n’?
(A) n nr E (B) n nr U (C) n
n
U
E (D) n
n
r
E
Solution (BC): 2
4 0n
r r .Z
n 0Z
U U .n
2
n 0 2
ZE E .
n
Thus; terms which are proportional to ‘n’ are
(i) n nr .U (ii) n
n
U
E
2. Select the correct statement:
(A) A process of expansion of an ideal gas which is isothermal as well as adiabatic must be a free-
expansion work (B) An adiabatic process in which no work is performed is an example of a constant energy process
(C) A reversible isothermal process can also be adiabatic only at T 0
(D) An irreversible isothermal process can also be adiabatic at T 0
Solution (ABCD):
(A) An isothermal free expansion is also adiabatic and vice versa.
(B) U Q W; for an adiabatic process in which no work is performed; i.e Q=0; W=0; thus;
U 0
(C) For rev isothermal process; 2
1
VU 0; W nRTln .
V
Thus; 2
1
VQ W nRTln
V
thus; at T 0 W 0; thus Q 0
(D) For are irreversible isothermal process extQ W P dV for it to become adiabatic also;
extP 0 ; thus at any T.
3. Which statements are correct about the two first order reactions if made with same initial
concentrations?
M Product
N product
(A) 1/2t for M is more than 1/2t of N
(B) 1/2t for M is less than 1/2t of N
(C) Rate constant for M is more than rate constant for N
(D) Rate constant for M is less than rate constant for N
Br + Mg Ether BrMg O
3(ii) H O
(i)2 2Ph CH CH OH
(A) (B)
SOCl AlCl2 32 2 2 2 (EAS)
Ph CH CH COOH Ph CH CH C Cl
O
(E) (F)
O
4LiAlH
OH
2 4H SO /
(G) (H) (I)
24
24
Solution (AD):
It is clear from graph that; 1/2 1/2t of M t of N
Again; 1/20.693
tK
thus; K for M<K for N.
4. Select species in which d-orbitals are used in hybridization and also involved in bond formation.
(A) 4OsF (B) 2 2XeO F (C) 3XeO (D) 6PCl
Solution (AB): 3
4OsF sp d hybridization and one p bond
33 2XeO F sp d hybridization and three p d bonds
5. Compound P: 2 4 2 2[Cr(H O) Cl ]Br 2H O
Q: 2 4 2 2[Cr(H O) Br ]Cl 2H O
R: 2 5 2 2[Cr(H O) Cl]Br 1H O
S: 2 6 3[Cr(H O) ]Cl
Incorrect statement(s) are:
(A) P and Q are ionization isomers
(B) P and R are hydrated isomers
(C) Given volume of solution of most conducting solution will require rd1/3 volume of equimolar
3AgNO solution for complete precipitation
(D) 2.0 grams of R prepared in 2L will show higher conduction compared with 2.0 gm of Q in 2L
solution
Solution (ABC):
Given vol. of must conducting solution requires thrice vol. of 3AgNO
6. Which of the following statements are not correct?
(A) Copper is extracted by self reduction method
(B) Cast iron is the purest form of iron
(C) The composition of malachite ore is 2 3Cu(OH) .2CuCO
(D) Cupellation process is used for the refining of Ag and Au
Solution (BC):
Conceptual
7. 2 4 2 42MX H SO (conc.) 2HX M SO (M=metal)
Above reaction is correct if ‘X’ is:
(A) F (B) Cl (C) 3NO
(D) I
Solution (ABC):
F and Cl salts produce HF and HCl respectively
I and Br salts produce 2I and 2Br respectively.
8. Which of the following reactions give diastereomeric products?
(A) (B)
(C) (D)
25
25
3CH
3CH Et
m CPBA
3CH
3CH Et
O
+
3CH
3CH Et
O
3CH
2 4Br /CCl
BrBr
3CH
+
BrBr
3CH
Solution (BC):
(B)
(C)
SECTION 4:
1. The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms speed of
B. If two samples of A and B contain same of molecule, the ratio of pressure of gas samples of A and B
in separate containers of equal volume is ……
Solution (8):
A BM 2M
rms rmsU (A) 2U (B)
A B A B A2
A B A B B
3RT 3RT T T T4 8
M M M M T
A AA
n .RTP
V B B A A
BB B
n .RT P TP 8
V P T
2. FeO crystallizes in NaCl type lattice. The crystal is however non-stoichiometric as 0.96Fe O and
deficient in iron. Some cation sites are vacant and some contain 3Fe
so that it becomes electrically
neutral. The % of cation sites vacant are……….
Solution (4):
For 1 mole Fe 4 octahedral voids
Thus for 0.96 mole of Fe 4 0.96 OV are occupied.
thus; % vacant sites4 0.96 4
100 4%4
3. If o o o1 2 3E ,E and E are standard oxidation potentials for
2 2 3 3Fe|Fe ,Fe |Fe and Fe|Fe , then
o oo 2 13
E 2EE
n
. The value of n is……
Solution (3):
2Fe Fe 2e ; o o1 1G 2FE …….(i)
2 3Fe Fe e ; o o2 2G 1 FE …….(ii)
3Fe Fe 3e ; o o3 3G 3 F E …….(iii)
Three;
o oo o o o 1 21 2 3 3
2E EG G G E
3
thus; n 3
4. Out of given aq. ion(s), find no. of ions which produce ppt. with dil. HCl + 2H S :
3 2 2 2 2 3 2Cr ,Zn ,Cd ,Hg ,Fe ,Bi ,Pb
Solution (4):
II group cations produce ppt.
26
26
5. In the compound 2 4 5 4 2Na [B O (OH) ] 8H O, if the
(i) number of B O B bonds is ‘x’
(ii) number of B B bonds is ‘y’
(iii) number o 2sp hybridized B atoms is ‘z’
Calculate the value of x y z .
Solution (7):
6. Identify number of compound from following. Which liberate 2CO on reaction with 3NaHCO .
Solution (5):
7. Identify number of carbohydrate that have L-configuration, from following:
Solution (5):
8. Identify number of substrate those can give NS 1 and NS 2 reaction both:
Solution (6):
27
27
MATHEMATICS (PAPER 2) SECTION 1:
PARAGRAPH FOR NEXT TWO QUESTIONS:
Let f x be non-positive continuous function and x
0
f x f t dt x 0 & f x f x where c 0 and
let g : 0, R be a function such that
dg x
g x x 0dx
and g 0 0
1. The total number of roots of the equation f x g x is
(A) (B) 1 (C) 2 (D) 0
Solution (B):
f x 0 and F x f x f x cF x
F x cF x 0 cx cxe F x ce F x 0
cxde F x 0
dx
cxe F x is an increasing function.
c 0cxe F x e F 0 cxe F x 0
f x 0 , (As f x cF x and c is positive) f x 0
Also,
cx dg xe g x x 0
dx
x xd g xe e g x 0
dx
xe g x is a decreasing function
0xe g x e g 0
g x 0 as g 0 0
Thus, f x g x has one solution x 0
2. The number of solutions of the equation 2x x 6 f x g x is/are
(A) 2 (B) 1 (C) 0 (D) 3
Solution (C):
2x x 6 f x g x 2x x 6 g x no solution
PARAGRAPH FOR NEXT TWO QUESTIONS:
The figure shown below, indicates positions of the ships A & B at t 0 . The ship A sails at 2km/hr due
east and ship B sails at 1km/hr along a course that forms an angle of o60 with the course of ship A.
3. The distance ‘d’ of separation of the ships as a function of time t is
(A) 23 t t 1 (B) 23 t 3t 5 (C) 23 t t 1 (D) 2t t 1
Solution (C):
4. What was the distance between the ships 2 hours before the initial moment shown in figure? (A) 1 (B) 3 (C) 4 (D) 6
Solution (B)
2d 3 2 2 1 3km.
28
28
Solution to question number (3 & 4):
Let at time t hrs the ship A is at /A and the ship B is at /B . Hence /PA 1 2t km and
/PB 2 t km. Let us take the line /PA as x-axis and P as origin. Hence coordinates of /A are
1 2t ,0 and that of /B are o o2 t cos60 , 2 t sin60 t t 3
1 , 32 2
.
22
t t 3d 1 1 2t 3 0
2 2
2 23t t
3 12 2
2 29t t3 1 t
4 4
2
212t 12t 123 t t 1
4
…(i)
2 22 1 1 1
3 t 2. .t 12 2 2
21 3
3 t2 4
21 9
3 t2 4
…(ii)
Clearly d is minimum if 1
t 02
, that is 1
t2
. Hence 1
hrs2
before the initial moment the ships were
minimum distance apart.
SECTION 2:
1. The perpendicular bisector x y 2 0 and x y 1 0 of sides AB & AC of a triangle ABC intersect,
them at 1, 1 & 2,1 respectively. If the middle points of side BC is P, then distance of this point
from the orthocenter of the triangle ABC is
(A) 85 (B) 41 (C) 13 (D) 7
Solution (C):
Since perpendicular bisector of AB, 4 : x y 2 0 and perpendicular bisector of AC, 2L : x y 1 0
are perpendicular to each other, therefore oA 90 .
A is the orthocenter & P will be the circumcenter. Hence the correct figure will be as shown below:
29
29
Solving 1 2L & L :
1 3P ,
2 2
Since ADPE is a rectangle, therefore PA DE
2 2PA DE PD PE
2 2 2
2
1 3 11 1 2
2 2 2PA
31
2
1 1 25 25
PA4 4 4 4
52
134
2. Let
0 1 2 n
n 23 4L 2C C C ... .C
2 3 n 1
(where n
r rC C ). If nn
LS lim
2 , then which of the following
is/are incorrect?
(A) S is prime (B) S is composite (C) S is even (D) S is odd
Solution (ABC): r n
nr
r 0
r 2L C
r 1
r nn
r
r 0
1 n 1L C r 2
n 1 r 1
n
n 1r 1
r 0
1L C r 2
n 1
n nn 1 n 1
r 1 r 1
r 0 r 0
1L r 1 C C
n 1
n n
n 1 r n 1r 1 r 1
r 1 r 0
1L r 1 C .1 C
n 1
n n 11
L n 1 2 2 1n 1
n nn n
L 1 1lim lim n 1 2
n 12 2
n nn n
L 2 1S lim lim 1 1
n 12 2 n 1
3. If the roots of the equation 4 3 2z z 36 15i z mz 0 are the vertices of a square then m
can be equal
(A) 35 45i (B) 35 45i (C) 35 45i (D) 35 45i
Solution (AB):
4 3 2z z 36 15i z mz 0
Clearly one of the roots of above equation is z 0 .
Since roots are vertices of square, therefore, let other three vertices be 1 1 1 1z ,iz ,z iz , (rotation of
complex number)
1 1 1 10 z iz z iz
12z 1 i …(i)
1 1 1 1 1 1 10 z 0 z iz 0 iz z z iz 1 1 1 1 1z iz iz z iz 36 15i
2 2 2 2 21 1 1 1 1z iz iz z iz 36 15i
213iz 36 15i 2
1z 5 12i 113 5 13 5
z i2 2
1z 9 i 4 1z 3 2i
30
30
CASE I: 1z 3 2i 1z 3 2i
From (i):
2 3 2i 1 i
2 3 2 i 2 3 2 1 5i 2 10i
1 1 1 1z z iz iz m 31z 1 i m
3
m 3 2i 1 i m 37 55i m 35 45i
CASE II: 1z 3 2i 1z 3 2i
From (i)
2 3 2i 1 i 2 10i
1 1 1 1z z iz iz m 31z 1 i m
3
m 3 2i 1 i m 37 55i m 35 45i m 35 45i
4. Let area of triangle formed by any line through point 1,3 and coordinate axes in xy-plane is A, then
(A) If A 8 , then number of such lines is 4 (B) If A 4 , then number of such lines is 4
(C) If A 6 , then number of such lines is 3 (D) If A 0 , then number of such lines is 1
Solution (AC):
Equation of line passing through 1,3 & having slope m is:
y 3 m x 1
For x 0 : y 3 m
For y 0 : 3
x 1m
Therefore area of triangle formed by the above line & coordinate axis is:
1 3
A 3 m 12 m
1 3 m
A 3 m2 m
3 m 1 3 m1
A2 m
21 1
A 3 m2 m
23 m1
A2 m
Verifying option (A):
A 8
23 m1
82 m
2
3 m16
m
2
3 m16
m
29 m 6m 16m 2m 6m 16m 9 0
2 2m 22m 9 0 or m 10m 9 0
Hence 4 values of ‘m’ are possible, since both equations have real & distinct roots. Hence option (A) is
correct.
Verifying option (B):
A 4
23 m1
42 m
29 m 6m 8m 2m 6m 8m 9 0
2 2m 14m 9 0 or m 2m 9 0
Hence 2 values of m are possible, since second quadratic have complex roots. Hence option (B) is not correct.
Verifying option (C):
A 6
23 m1
62 m
29 m 6m 12m
2m 6m 12m 9 0 2 2m 18m 9 0 or m 6m 9 0
22m 18m 9 0 or m 3 0
31
31
Hence 3 values of m are possible, since second quadratic have equal roots. Hence option (C) is correct.
Verifying option (D):
A 0
23 m1
02 m
m 3
Also for m 0 or m area of is 0.
Hence number of such lines is 3. Hence option (D) is not correct.
5. Two vertices of an equilateral triangle are 0,0 & 0,2 and the third vertex lies in 2nd quadrant. Then
(A) The equation of circumcircle of the triangle is 2
21 4x y 1
33
(B) The equation of incircle of the triangle is 2
21 1x y 1
33
(C) The area of the square that circumscribe the circum circle of the triangle is 16
3.
(D) The area of the square inscribed in the circumcircle of the triangle is 3
16.
Solution (ABC):
Coordinates of point B are:
o2cos30,2sin30 3,1
Hence centroid G of the triangle OAB is:
0 0 3 0 2 1 1G , ,1
3 3 3
Since the triangle is equilateral therefore all its centres coincide.
Circumradius 2
21 2R GA 2 1
3 3
.
Hence equation of circum circle is:
2
21 4x y 1
33
.
Inradius 2
21 1r GM 1 1
3 3
Hence equation of incircle is:
2
21 1x y 1
33
.
If a square circumscribes a circle of radius R then area of the square is 2
2 2 162R 2
33
.
32
32
6. Which of the following statements are true?
(A) If A is an invertible 3 3 matrix and B is a 3 4 matrix, then 1A B is defined.
(B) It is never true that A B,A B , and AB are all defined.
(C) Every matrix none of whose entries are zero is invertible.
(D) Every invertible matrix is square and has no two rows the same.
Solution (AB):
7. Lines 1L ax by c 0 and 2L x my n 0 intersect at the point P and make an angle with
each other. Consider the line L, different from 2L and passing through the point P and making the
same angle with 1L . Then
(A) The equation of line L is 2 22 a bm ax by c a b x my n 0
(B) The equation of line L is 2 22 a bm ax by c a b x my n 0
(C) The line L is reflection of the line 2L about the line 1L
(D) The line L is reflection of the line 1L about the line 2L
Solution (AC):
Since L & 2L are equally inclined to 1L and L, 1 2L ,L are concurrent therefore 1L is one of the angle
bisectors of L and 2L . Hence L is reflection of 2L about the line 1L . Also, if 1 1A x ,y is any arbitrary
point on 1L then perpendicular distance of the line 2L from A is equal to perpendicular distance of the
line L from A.
1 1 1 11 1
2 2 2 2
ax by c x my nx my n
m a b m
But 1 1ax by c 0 .
1 11 1
2 2 2 2
x my nx my n
m a b m
2 2 2 2
1
m a b m
2 2 2 2 2a b m m
2 2 2 2 2 2 2 2a b m 2 a bm m
2 2a b
2 a bm
.
2 2a b
L ax by c x my n 02 a bm
2 22 a bm ax by c a b x my n 0
8. Identify the statements which are True.
(A) the equation of the director circle of the ellipse, 2 25x 9y 45 is 2 2x y 14.
(B) the sum of the focal distances of the point 0,6 on the ellipse 2 2x y
125 36
is 10.
(C) The point of intersection of any tangent to a parabola and the perpendicular to it from the focus lies
on the tangent at the vertex.
(D) P and Q are the points with eccentric angles & on the ellipse 2 2
2 2
x y1
a b , then the area of
the triangle OPQ is independent of , where O is origin.
33
33
Solution (ACD):
Let us verify each option individually:
(A) 2 25x 9y 45 2 2x y
19 5
Hence equation of director circle is: 2 2x y 9 5 2 2x y 14.
Hence option (A) is correct.
(B) Sum of focal distances of any point on any ellipse is equal to length of major axis of the ellipse. For
the given ellipse length of major axis is 2 6 12. Hence option (B) is false.
(C) Option (C) is correct (property of parabola).
NOTE: If SM be perpendicular to tangent at any point P, then M lies on the tangent at the vertex
and 2SM OS.SP , where S is the focus and O is the vertex of the parabola
Proof:
Let 2y 4ax be the given parabola. Tangent at 1 1P x ,y is S a,0 …(i)
The above tangent intersects y-axis at i.e.
21
1
1
y2a
4a y0, 0,
y 2
. The equation of the line passing
through S a,0 and perpendicular to the tangent is:
1yy 0 x a
2a
1y
y x a2a
…(ii)
The line (ii) intersects y-axis at 1y0,
2
, which is same as co-ordinates of M. Hence (i) and (ii)
intersect on y-axis i.e. the tangent at vertex of the parabola.
Also in triangle PMS SM
sinSP
In triangle SOM OS
sinSM
OS SM
SM SP 2SM OS.SP
(D) P acos ,bsin
Q acos ,bsin
O 0,0
Hence area of triangle OPQ is given by:
a cos bsin 11
a cos bsin 12
0 0 1
34
34
1
abcos .sin abcos .sin2
1 1
ab sin ab sin2 2
Hence option (D) is correct.
SECTION 4:
1. If /f x g x and /g x f x for all x and /f 2 4 f 2 then the value of 2 2f 19 g 19 27 is
Solution (5):
/ /f x g x ;g x f x
/f 2 f 2 4 g 2 4
Let h(x) 2 2f x g x 27
/ / /h x 2f x .f x 2g x .g x
/ /2f x .f x 2f x . f x 0
h x constant
2 2h 19 h 2 f 2 g 2 27 32 27 5
2. If 6 xf x 7 x and 1g x f x then the value of 7 /ln 7 e g 2 is
Solution (1):
1g x f x
1f g x f f x x
df g x
1dx
/ /f g x .g x 1
/
/
1g x
f g x
/
/
1g 2
f g 2
Let g 2 t
1f 2 t 1f f 2 f t
2 f t f t 2 6 t7 t 2 6 t7 t 2
By trial t 5 satisfies the above equation. Also 6 ty 7 is a decreasing function of t while y t 2 is
an increasing function of t. Hence the graph of 6 ty 7 and y t 2 can intersect at most once. Hence
t 5 is the only solution.
g 2 5
/
/ /
1 1g 2
f g 2 f 5
6 x
x 5
1
7 ln7 1
7
1 1
7ln7 1 ln7 lne 7
1
ln 7 e
7 /ln 7 e g 2 1
3. A variable plane at a distance of 1unit from the origin cuts the co-ordinate axes at A, B and C. If the
centroid G x,y,z of the triangle ABC satisfies the relation 2 2 2
1 1 1k
x y z , then the value of k is
35
35
Solution (9):
Let the plane meets the coordinate axes at A a,0,0 ,B 0,b,0 ,C 0,0,c . Hence the equation of the
plane ABC is x y z
1a b c . Also centroid G of the triangle ABC is
a b c, ,
3 3 3
.
a b c
x , y , z3 3 3
a 3x, b 3y, c 3z .
As per question the distance of the plane x y z
1a b c from origin is 1.
2 2
11
1 1 1
a b c
2 2 2
1 1 11
a b c
2 2 2
1 1 19
x y z .
4. Let 2A,G and H be the roots of cubic equation 3 2x px qx r 0 which are in GP, where p, q are
integers and A, G, H are respectively AM, GM, HM of two positive numbers. If p, q 100, 100 , then
let number of all possible ordered triplets p,q,r is N. Value of N 90 is
Solution (7):
A, 2G ,H are roots of the equation 3 2x px qx r 0 and roots are also in GP.
2
2G A H …(i)
4G A H 4 2 2G G , G AH
2 2G G 1 0 2 2G 0 or G 1 G 1, 0, 1
Since G is geometric mean of two positive numbers.
G 1
AH 1 , (using (i))
1
HA
Hence roots of the equation are:
1A,1,
A
31 r
A 1 1A 1
1 r r 1
1A 1 p
A …(ii)
1
p A 1 2 1, AM GMA
p 3 p 3 p 99, 98,..., 4, 3
1 1A 1 1 A q
A A
1
A 1 qA
…(iii)
1
q A 1A
q 2 1, AM GM q 3 q 3,4,5,...,99
From (ii) & (iii): p q 0
Now since p q 0 , for every value of p, value of q is fixed, possible ordered triplets (p, q, r) are:
99,99, 1 , 98,98,1 ,..., 3,3,1 .
Number of possible ordered triplets are 97.
N 97 N 90 7
36
36
5. If & are roots of the equation
5 5
10 21 1
0 01 1 x 5x 202 2
1 25 401 2 0 1 x 2
1 12 2
then find the
value of 2 2 is
Solution (6):
Let
10
2A
11
2
& 1 1
B2 0
1 10 1 2 0 1 0
2 2AB
1 11 1 2 1 1 0
2 2
1 0I
0 1
AB BA I 5 5
101 1
0 01 12 2
1 2 0 11 1
2 2
5 10 5A B A
10 times
AAAAA BBBB.......BAAAAA
8 times
AAAAI BBB.......B IAAAA , AB BA I
8 times
AAAA BBB.......B AAAA , AI IA A
6 times
AAAI BBB.......B IAAA , AB BA I
6 times
AAA BBB.......B AAA , AI IA A
2 times
A BB A I I I
Given
5 5
10 21 1
0 01 1 x 5x 202 2
1 25 401 2 0 1 x 2
1 12 2
2x 5x 20
1 25 I 40x 2
2x 5x 20
1 25 40x 2
, AI IA A
21 x 5x 20 25 x 2 40
2x 5x 20 25x 50 40
2x 20x 30 0
As , root of the above equation:
2x 20x 30 1 x x
37
37
For x 2 :
2
2 20 2 30 2 2
4 40 30 2 2
6 2 2 6 2 2
6. A function f x satisfies f 1 3600 and 2f 1 f 2 ... f n n f n for all positive integers n 1 .
Find the value of f 9
10 is equal to
Solution (8):
2f 1 f 2 f 3 ... f n 1 f n n f n …(i)
Replacing n by n 1 :
2
f 1 f 2 ... f n 1 n 1 f n 1 …(ii)
Subtracting (ii) from (i):
22f n n f n n 1 f n 1 , n 2
221 n f n n 1 f n 1
2
2
n 1 f n 1f n
n 1
n 1f n f n 1
n 1
…(iii)
Replacing n by n 1 in (iii):
n 2f n 1 f n 2
n
Replacing n by n 2 in (iii):
n 3f n 2 f n 3
n 1
etc.
n 1f n f n 1
n 1
n 1 n 2
. f n 2n 1 n
n 1 n 2 n 3
. . f n 3n 1 n n 1
……………………….
……………………….
n 1 n 2 n 3 3 2 1
. . .... . . .f 1n 1 n n 1 5 4 3
2 1
f 1n 1 n
2
3600n 1 n
2 3600
f 99 1 9
2 360080
10 9
f 98
10 .
7. The least possible value of a for which
3 2
3 2
x 6x 11x 6 a0
30x x 10x 8
does not have a real solution is
Solution (5):
Let a
f x 030
38
38
Where 3 2
3 2
x 6x 11x 6f x
x x 10x 8
x 1 x 2 x 3
x 1 x 2 x 4
x 3
,x 1,2, 4x 4
Range of 2 1
f x R 1, ,5 6
So, the equation does not have solution if
a 2 11, ,
30 5 6
a 30,12,5
8. Let
sin xdF x e
dx x
, x 0 if
4 3sin x
1
3e dx
x
F k F 1 , then the possible value of k is
Solution (8):
In
4 3sin x
1
3e dx
x, put
3x t , so that 33x dx dt .
Also, when x 1,t 1 and when x 4,t 64 .
Therefore,
4 643sinx dx sint2
1 1
3 3 dte e
x x 3x
64 64
sint
1 1
1 de dt F t dt F 64 F 1
t dt
Hence, k 64