JEE MAINS Solved Paper 2007

Upload
chithrasajeev 
Category
Documents

view
29 
download
0
description
Transcript of JEE MAINS Solved Paper 2007
Instructions
1. This test consists of 120 questions.
2. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 40 questions in each part of equal weightage. Each question is allotted 4 marks for correct
response.
3. Candidates will be awarded marks as stated above for correct response of each question. 1/4 mark will
be deducted for indicating incorrect response of each question. No deduction from the total score will
be made if no response is indicated for an item in the answer sheet.
4. There is only one correct response for each question. Filling up more than one response in any
question will be treated as wrong response and marks for wrong response will be deducted according
as per instructions.
Physics1. The displacement of an object attached
to a spring and executing simpleharmonic motion is given by x = ´ 2 10 2
cos pt metre. The time at which themaximum speed first occurs is
(a) 0.5 s (b) 0.75 s(c) 0.125 s (d) 0.25 s
2. In an AC circuit, the voltage applied is
E E t= 0 sin w . The resulting current in the
circuit is I I t= æèç
öø÷0 2
sin w p. The power
consumption in the circuit is given by
(a) PE I= 0 0
2(b) P = zero
(c) PE I= 0 0
2(d) P E I= 2 0 0
3. An electric charge 10 3 mC is placed at the
origin (0, 0) of xycoordinate system. Twopoints A and B are situated at ( , )2 2 and
(2, 0) respectively. The potential differencebetween the points A and B will be
(a) 9 V (b) zero(c) 2 V (d) 4.5 V
4. A battery is used to charge a parallel platecapacitor till the potential differencebetween the plates becomes equal to theelectromotive force of the battery. The ratioof the energy stored in the capacitor andthe work done by the battery will be
(a) 1 (b) 2
(c)14
(d)12
Solved Paper 2007
All India Engineering Entrance Examination
5. An ideal coil of 10 H is connected in serieswith a resistance of 5 W and a battery of5 V. After 2 s, the connection is made, thecurrent flowing (in ampere) in the circuitis
(a) ( )1  e (b) e
(c) e1 (d) ( )1 1 e
6. A long straight wire of radius a carries asteady current i. The current is uniformlydistributed across its crosssection. The
ratio of the magnetic field ata
2and 2 a is
(a)14
(b) 4 (c) 1 (d)12
7. A current I flows along the length of aninfinitely long, straight, thin walled pipe.Then,(a) the magnetic field is zero only on the axis of the
pipe(b) the magnetic field is different at different points
inside the pipe(c) the magnetic field at any point inside the pipe
is zero(d) the magnetic field at all points inside the pipe
is the same, but not zero
8. If MO is the mass of an oxygen isotope
817O , Mp and Mn are the masses of a
proton and a neutron respectively, thenuclear binding energy of the isotope is
(a) ( )M M cpO  8 2 (b) ( )M M M cp nO  8 9 2
(c) M cO2 (d) ( )M M cnO  17 2
9. In gamma ray emission from a nucleus
(a) both the neutron number and the protonnumber change
(b) there is no change in the proton number andthe neutron number
(c) only the neutron number changes(d) only the proton number changes
10. If in a pn junction diode, a square inputsignal of 10 V is applied as shown
Then, the output signal across RL will be
11. Photon of frequency n has a momentumassociated with it. If c is the velocity oflight, the momentum is
(a) n / c (b) hn c (c) h cn/ 2 (d) h cn/
12. The velocity of a particle is
v v gt ft= + +02 . If its position is x = 0 at
t = 0, then its displacement after unit time( )t = 1 is
(a) v g f0 2 3+ + (b) v g f0 2 3+ +/ /(c) v g f0 + + (d) v g f0 2+ +/
13. For the given uniformsquare lamina ABCD,whose centre is O
(a) 2 I IAC EF=
(b) I IAD EF= 4
(c) I IAC EF=(d) I IAC EF= 2
14. A point mass oscillates along the xaxisaccording to the law x x= 0 cos ( / )w pt  4 .If the acceleration of the particle is writtenas a A t= +cos( ),w d then
(a) A x= = 0 4, /d p (b) A x= =02 4w d p, /
(c) A x= = 02 4w d p, / (d) A x= =0
2 3 4w d p, /
15. Charges are placed onthe vertices of a squareas shown. Let E be theelectric field and V bethe potential at thecentre. If the charges onA and B are interchanged with those on D
and C respectively, then
(a) E remains unchanged, V changes(b) both E and V change(c) E and V remain unchanged(d) E changes, V remains unchanged
176 JEE Main Solved Papers
–5 V
(a) (b)
(c) (d)
+5 V
10 V
–10 V
5 V
– 5 V
RL
D C
O
BAE
F
A B
CD
q q
–q –q
16. The halflife period of a radioactiveelement X is same as the mean life time ofanother radioactive element Y. Initially,they have the same number of atoms.Then,
(a) X will decay faster than Y
(b) Y will decay faster than X
(c) Y and X have same decay rate initially(d) X and Y decay at same rate always
17. A Carnot engine, having an efficiency ofh = 1 10/ as heat engine, is used as arefrigerator. If the work done on thesystem is 10 J, the amount of energyabsorbed from the reservoir at lowertemperature is
(a) 99 J (b) 90 J (c) 1 J (d) 100 J
18. Carbon, silicon and germanium have fourvalence electrons each. At roomtemperature, which one of the followingstatements is most appropriate?
(a) The number of free conduction electrons issignificant in C but small in Si and Ge
(b) The number of free conduction electrons isnegligibly small in all the three
(c) The number of free electrons for conduction issignificant in all the three
(d) The number of free electrons for conduction issignificant only in Si and Ge but small in C
19. A charged particle with charge q enters aregion of constant, uniform and mutuallyorthogonal fields E and B with a velocity v
perpendicular to both E and B and comesout without any change in magnitude ordirection of v. Then,
(a) v E B= ´ / B2 (b) v B E= ´ / B2
(c) v E B= ´ / E2 (d) v B E= ´ / E2
20. The potential at a point x (measured inmm) due to some charges situated on thexaxis is given by V x x( ) / ( )= 20 42 volt.
The electric field E at x = 4 mm is given by
(a)53
V/mm and in the –ve x direction
(b)53
V/mm and in the +ve x direction
(c)109
V/mm and in the –ve x direction
(d)109
V/mm and in the +ve x direction
21. Which of the following transitions inhydrogen atoms emit photons of highestfrequency?
(a) n = 2 to n = 6 (b) n = 6 to n = 2
(c) n = 2 to n = 1 (d) n = 1to n = 2
22. A block of mass m is connected to anotherblock of mass M by a spring (massless) ofspring constant k. The blocks are kept ona smooth horizontal plane. Initially theblocks are at rest and the spring isunstretched. Then, a constant force F
starts acting on the block of mass M to pullit. Find the force on the block of mass m.
(a)mFM
(b)( )M m F
m
+
(c)mF
m M( )+(d)
MFm M( )+
23. Two lenses of power –15D and +5D are incontact with each other. The focal lengthof the combination is
(a) –20 cm (b) –10 cm
(c) +20 cm (d) +10 cm
24. One end of a thermally insulated rod iskept at a temperature T1 and the other atT2 . The rod is composed of two sections oflengths l1 and l2 and thermalconductivities K1 and K2 ,respectively. Thetemperature at the interface of the twosections is
(a) ( )/( )K T K T K K2 2 1 1 1 2 1 1 2 2l l l l+ +
(b) ( )/( )K T K T K K2 1 1 1 2 2 2 1 1 2l l l l+ +
(c) ( )/( )K T K T K K1 2 1 2 1 2 1 2 2 1l l l l+ +
(d) ( )/( )K T K T K K1 1 1 2 2 2 1 1 2 2l l l l+ +
25. A sound absorber attenuates the soundlevel by 20 dB. The intensity decreases bya factor of
(a) 1000 (b) 10000(c) 10 (d) 100
AIEEE Solved Paper 2007 177
T1 T2
K1 K2
l1 l2
26. If CP and CV denote the specific heats of
nitrogen per unit mass at constantpressure and constant volumerespectively, then
(a) C C Rp V = /28 (b) C C Rp V = /14
(c) C C Rp V = (d) C C Rp V = 28
27. A charged particle moves through amagnetic field perpendicular to itsdirection. Then,
(a) the momentum changes but the kinetic energyis constant
(b) both momentum and kinetic energy of theparticle are not constant
(c) both momentum and kinetic energy of theparticle are constant
(d) kinetic energy changes but the momentum isconstant
28. Two identical conducting wires AOB andCOD are placed at right angles to eachother. The wire AOB carries an electriccurrent I1 and COD carries a current I2 .The magnetic field on a point lying at adistance d from O, in a directionperpendicular to the plane of the wiresAOB and COD, will be given by
(a)m
p0 1 2
1 2
2
I I
d
+æèç
öø÷
/
(b)mp
012
22 1 2
2 dI I( ) /+
(c)mp
01 22 dI I( )+ (d)
mp
012
22
2 dI I( )+
29. The resistance of a wire is 5 W at 50°C and6 W at 100°C. The resistance of the wire at0°C will be
(a) 2 W (b) 1W
(c) 4W (d) 3W
30. A parallel plate condenser with adielectric of dielectric constant K betweenthe plates has a capacity C and is chargedto a potential V volts. The dielectric slab isslowly removed from between the platesand then reinserted. The net work done bythe system in this process is
(a)12
1 2( )K CV (b) CV K K2 1( )/
(c) ( )K CV 1 2 (d) zero
31. If gE and gM are the accelerations due to
gravity on the surfaces of the earth andthe moon respectively and if Millikan’s oildrop experiment could be performed onthe two surfaces, one will find the ratioelectronic charge on the moon
electronic charge on the earthto be
(a) 1 (b) zero (c) g gE M/ (d) g gM E/
32. A circular disc of radius R is removed froma bigger circular disc of radius 2R, suchthat the circumference of the discscoincide. The centre of mass of the new
disc isaR
from the centre of the bigger
disc. The value of a is
(a)13
(b)12
(c)16
(d)14
33. A round uniform body of radius R, mass M
and moment of inertia I, rolls down(without slipping) an inclined planemaking an angle q with the horizontal.Then, its acceleration is
(a)g
I MR
sin
/
q1 2+
(b)g
MR I
sin
/
q1 2+
(c)g
I MR
sin
/
q1 2
(d)g
MR I
sin
/
q1 2
34. Angular momentum of the particlerotating with a central force is constantdue to
(a) constant force(b) constant linear momentum(c) zero torque(d) constant torque
35. A 2 kg block slides on a horizontal floorwith a speed of 4 m/s. It strikes auncompressed spring and compresses ittill the block is motionless. The kineticfriction force is 15 N and spring constantis 10000 N/m. The spring compresses by
(a) 9.34 cm (b) 2.5 cm
(c) 11.0 cm (d) 8.5 cm
36. A particle is projected at 60° to thehorizontal with a kinetic energy K. Thekinetic energy at the highest point is
(a) K (b) zero (c) K/4 (d) K/2
178 JEE Main Solved Papers
37. In a Young’s doubleslit experiment, theintensity at a point where the path
difference isl6
(l being the wavelength of
the light used), is I. If I0 denotes themaximum intensity, I I/ 0 is equal to
(a)1
2(b)
32
(c)12
(d)34
38. Two springs, of force constants k1 and k2 ,
are connected to a mass m as shown. Thefrequency of oscillation of the mass is f. Ifboth k1 and k2 are made four times theiroriginal values, the frequency ofoscillation becomes
(a) f/2 (b) f/4 (c) 4f (d) 2 f
39. When a system is taken from state i tostate f along the path iaf, it is found thatQ = 50 cal and W = 20 cal. Along the pathibf Q, = 36 cal. W along the path ibf is
(a) 6 cal (b) 16 cal(c) 66 cal (d) 14 cal
40. A particle of mass m executes simpleharmonic motion with amplitude a andfrequency n. The average kinetic energyduring its motion from the position ofequilibrium to the end is
(a) p n2 2 2ma (b)14
2 2ma n
(c) 4 2 2 2p nma (d) 2 2 2 2p nma
Chemistry41. The energies of activation for forward and
reverse reactions for A B AB2 2 2+ s
are 180 kJ mol 1 and 200 kJ mol 1
respectively. The presence of a catalystlowers the activation energy of both(forward and reverse) reactions by100 kJ mol1. The enthalpy change of the
reaction ( )A B AB2 2 2+ ¾® in thepresence of catalyst will be (in kJ mol 1 )
(a) 300 (b) 120 (c) 280 (d)  20
42. The cell, ZnZn (1M)Cu (1M)Cu2+ 2+
(E° =cell 1.10 V), was allowed to becompletely discharged at 298 K. Therelative concentration of Zn2+ to Cu2+
[Zn ]
[Cu ]
2
2
+
+
æèç
öø÷ is
(a) antilog (24.08) (b) 37.3(c) 1037 3. (d) 9 65 104. ´
43. The pKa of a weak acid (H A) is 4.5. The
pOH of an aqueous buffered solution ofHA in which 50% of the acid ionised is
(a) 4.5 (b) 2.5 (c) 9.5 (d) 7.0
44. Consider the reaction,
2 A B+ ¾® products
When concentration of B alone wasdoubled, the halflife did not change.When the concentration of A alone wasdoubled, the rate increased by two times.The unit of rate constant for this reactionis(a) L mol s1 1  (b) no unit(c) mol L s1 1  (d) s 1
45. Identify the incorrect statement amongthe following.
(a) dblock elements show irregular and erraticchemical properties among themselves
(b) La and Lu have partially filled d orbitals and noother partially filled orbitals
(c) The chemistry of various lanthanoids is verysimilar
(d) 4f and 5f orbitals are equally shielded
46. Which one of the following has a squareplanar geometry?
(At. no. Co = ,27 Ni 28,= Fe 26= , Pt 78)=(a) [CoCl ]4
2– (b) [FeCl ]42–
(c) [NiCl ]42– (d) [PtCl ]4
2–
AIEEE Solved Paper 2007 179
k1 k2m
i
a f
b
47. Which of the following molecules is expectedto rotate the plane of planepolarised light?
48. The secondary structure of a protein refers to
(a) ahelical backbone(b) hydrophobic interactions(c) sequence of aamino acids(d) fixed configuration of the polypeptide backbone
49. Which of the following reactions will yield, 2,2dibromopropane?
(a) CH C CH + 2HBr3 ¾ ºº ¾®(b) CH CH==CHBr + HBr3 ¾®(c) CH CH + 2HBrºº ¾®(d) CH CH==CH + HBr3 2¾ ¾®
50. In the chemical reaction,
CH CH NH + CHCl 3KOH3 2 2 3 + ¾®( ) ( )A B+ + 3 O2H , the compounds (A) and(B) are respectively
(a) C H CN2 5 and 3KCl
(b) CH CH CONH3 2 2 and 3KCl
(c) C H NC2 5 and K CO2 3
(d) C H NC2 5 and 3KCl
51. The reaction of toluene with Cl2in the presence ofFeCl3 gives predominantly
(a) benzoyl chloride(b) benzyl chloride(c) o  and pchlorotoluene(d) mchlorotoluene
52. Presence of a nitro group in a benzene ring
(a) activates the ring towards electrophilic substitution(b) renders the ring basic(c) deactivates the ring towards nucleophilic
substitution(d) deactivates the ring towards electrophilic
substitution
53. In which of the following ionisationprocesses, the bond order hasincreased and the magnetic behaviourhas changed?
(a) C C2 2+¾® (b) NO NO+¾®
(c) O O2 2+¾® (d) N N2 2
+¾®
54. The actinoids exhibit more number ofoxidation states in general than thelanthanoids. This is because
(a) the 5f orbitals are more buried than the
4f orbitals(b) there is a similarity between 4f and 5f
orbitals in their angular part of the wavefunction
(c) the actinoids are more reactive than thelanthanoids
(d) the 5f orbitals extend further from thenucleus than the 4f orbitals
55. Equal masses of methane and oxygen
are mixed in an empty container at
25°C. The fraction of the total pressure
exerted by oxygen is
(a)23
(b)13
273298
´
(c)13
(d)12
56. A 5.25% solution of a substance isisotonic with a 1.5% solution of urea(molar mass = 60 g mol 1 ) in the samesolvent. If the densities of both thesolutions are assumed to be equal to1.0 g cm 3 , the molar mass of thesubstance will be
(a) 90.0 g mol–1 (b) 115.0 g mol–1
(c) 105.0 g mol–1 (d) 210.0 g mol–1
57. Assuming that water vapour is anideal gas, the internal energy change( )DU when 1 mole of water isvapourised at 1 bar pressure and100°C, (Given : molar enthalpy ofvaporisation of water at 1 bar and 373K = 41kJ mol 1 and R = 8.3 J mol 1 K )1
will be
(a) 4.100 kJ mol 1 (b) 3.7904 kJ mol 1
(c) 37.904 kJ mol 1 (d) 41.00 kJ mol 1
180 JEE Main Solved Papers
CHO
H
CH OH2
HO(a)
SH
(b)
COOH
H
H
H N2
H N2
Ph
H
Ph
NH2
H(c) (d)
58. In a saturated solution of the sparinglysoluble strong electrolyte AgIO3
(Molecular mass = 283), the equilibriumwhich sets in, is
AgIO ( ) Ag ( ) IO ( )3 3s aq aqs+ +
if the solubility product constant, Ksp ofAgIO3 at a given temperature is1.0 ´ 10 8 , what is the mass of AgIO3
contained in 100 mL of its saturatedsolution?
(a) 28 3 10 2. ´  g
(b) 2 83 10 3. ´  g
(c) 10 10 7. ´  g
(d) 10 10 4. ´  g
59. A radioactive element gets spilled overthe floor of a room. Its halflife period is 30days. If the initial activity is ten times thepermissible value, after how many dayswill it be safe to enter the room?
(a) 1000 days(b) 300 days(c) 10 days(d) 100 days
60. Which one of the following conformationsof cyclohexane is chiral?
(a) Twist boat (b) Rigid(c) Chair (d) Boat
61. Which of the following is the correct orderof decreasing S 2N reactivity?(X a= halogen)
(a) R X R X R XCH C CH2 3 2> >
(b) R X R X R XCH CH C2 2> > 3
(c) R X R X R X3 2C CH CH2> >
(d) R X R X R X2 3 2CH C CH> >
62. In the following sequence of reactions
CH CH OH3 2
P + I Mg
Ether
HCHO2¾¾® ¾® ¾¾®A B
C D¾¾®H O2
the compound ‘D’ is
(a) butanal (b) nbutyl alcohol(c) npropyl alcohol (d) propanal
63. Which of the following sets of quantumnumbers represents the highest energy ofan atom?
(a) n l m s= = = = +3 1 1 1 2, , , /(b) n l m s= = = = +3 2 1 1 2, , , /(c) n l m s= = = = +4 0 0 1 2, , , /(d) n l m s= = = = +3 0 0 1 2, , , /
64. Which of the following hydrogen bonds isthe strongest?
(a) O H N¾ K
(b) F H F¾ K
(c) O H O¾ K
(d) O H F¾ K
65. In the reaction,
2Al( ) 6HCl( ) 2Al ( )3+s aq aq+ ¾®+ +6Cl ( ) 3H ( )2aq g
(a) 6 L HCl ( )aq is consumed for every 3 L H2( )gproduced
(b) 33.6 L H2 (g) is produced regardless oftemperature and pressure for every mole of Althat reacts
(c) 67.2 L H2( )g at STP is produced for every moleof Al that reacts
(d) 11.2 L H2( )g at STP is produced for every moleof HCl(aq) consumed
66. Regular use of which of the followingfertilisers increases the acidity of soil?
(a) Potassium nitrate(b) Urea(c) Superphosphate of lime(d) Ammonium sulphate
67. Identify the correct statement regarding aspontaneous process
(a) For a spontaneous process in an isolatedsystem, the change in entropy is positive
(b) Endothermic processes are neverspontaneous
(c) Exothermic processes are alwaysspontaneous
(d) Lowering of energy in the reaction process isthe only criterion for spontaneity
68. Which of the following nuclear reactionswill generate an isotope?
(a) Neutron particle emission(b) Positron emission(c) aparticle emission(d) bparticle emission
AIEEE Solved Paper 2007 181
69. The equivalent conductances of two
strong electrolytes at infinite dilution in
H O2 (where ions move freely through a
solution) at 25°C are given below
L°CH COONa2
391.0 S cm / equiv=
L°HCl2426.2 S cm / equiv=
What additional information/quantity oneneeds to calculate L° of an aqueoussolution of acetic acid?
(a) L° of NaCl(b) L° of CH COOK3
(c) The limiting equivalent conductance of
HH
+ ° +( )l(d) L° of chloroacetic acid (ClCH COOH)2
70. Which one of the following is thestrongest base in aqueous solution?
(a) Trimethylamine (b) Aniline(c) Dimethylamine (d) Methylamine
71. The compound formed as a result ofoxidation of ethyl benzene by KMnO4 is
(a) benzophenone (b) acetophenone(c) benzoic acid (d) benzyl alcohol
72. The IUPAC name of is
(a) 1, 1diethyl2, 2dimethylpentane(b) 4, 4dimethyl5, 5diethylpentane(c) 5, 5diethyl4, 4dimethylpentane(d) 3ethyl4, 4dimethylheptane
73. Which of the following species exhibitsthe diamagnetic behaviour?
(a) O22 (b) O2
+
(c) O2 (d) NO
74. The stability of dihalides of Si, Ge, Sn andPb increases steadily in the sequence
(a) Ge Si Sn Pb2 2 2 2X X X X< < <(b) Si Ge Pb Sn2 2 2 2X X X X< < <(c) Si Ge Sn Pb2 2 2 2X X X X< < <(d) Pb Sn Ge Si2 2 2 2X X X X< < <
75. Identify the incorrect statement amongthe following.
(a) Ozone reacts with SO2 to give SO3
(b) Silicon reacts with NaOH ( )aq in the presenceof air to give Na SiO2 3 and H O2
(c) Cl2 reacts with excess of NH3 to give N2 andHCl
(d) Br2 reacts with hot and strong NaOH solutionto give NaBr, NaBrO4 and H O2
76. The charge/size ratio of a cationdetermines its polarizing power. Whichone of the following sequences representsthe increasing order of the polarizingpower of the cationic species : K+ , Ca ,2+
Mg ,2+ Be2+?
(a) Mg < Be < K < Ca2+ 2+ + 2+
(b) Be < K < Ca < Mg2+ + 2+ 2+
(c) K < Ca < Mg < Be+ 2+ 2+ 2+
(d) Ca < Mg < Be < K2+ 2+ 2+ +
77. The density (in g mL )1 of a 3.60 M
sulphuric acid solution that is 29% H SO2 4
(molar mass = 98 gmol )1 by mass will be
(a) 1.64 (b) 1.88 (c) 1.22 (d) 1.45
78. The first and second dissociationconstants of an acid H2 A are 1.0 10–5´and 5.0 ´ 10 10 respectively. The overall
dissociation constant of the acid will be
(a) 5 0 10 5. ´  (b) 5 0 1015. ´(c) 5 0 10 15. ´  (d) 0 2 105. ´
79. A mixture of ethyl alcohol and propyl
alcohol has a vapour pressure of 290 mm
at 300 K. The vapour pressure of propyl
alcohol is 200 mm. If the mole fraction of
ethyl alcohol is 0.6, its vapour pressure (in
mm) at the same temperature will be
(a) 350 (b) 300 (c) 700 (d) 360
80. In conversion of limestone to lime,
CaCO CaO CO23( ) ( ) ( )s s g¾® +
the values of DH° and DS° are
+ 1791 1. kJ mol and 160.2 J/K respectively
at 298 K and 1 bar. Assuming that DH° and
DS° do not change with temperature,
temperature above which conversion of
limestone to lime will be spontaneous is
(a) 1008 K (b) 1200 K(c) 845 K (d) 1118 K
182 JEE Main Solved Papers
Mathematics81. In a geometric progression consisting of
positive terms, each term equals the sum ofthe next two terms. Then, the common ratioof this progression equals
(a)12
1 5( ) (b)12
5
(c) 5 (d)12
5 1( )
82. If sin5
cosec54 2
1 1 æèç
öø÷ + æ
èçöø÷ =x p
, then a value of
x is
(a) 1 (b) 3 (c) 4 (d) 5
83. In the binomial expansion of ( )a b n , n ³ 5,
the sum of 5th and 6th terms is zero, thena
b
equals
(a)5
4n (b)
65n 
(c)n  5
6(d)
n  45
84. The set S = , , , ..., 1 2 3 12 is to be partitioned
into three sets A B C, , of equal size. Thus,A B C SÈ È = ,A B B C A CÇ = Ç = Ç = f. The number ofways to partition S is
(a) 12!/3! ( !)4 3 (b) 12!/3!( !)3 4
(c) 12!/( !)4 3 (d) 12!/( !)3 4
85. The largest interval lying in æèç
öø÷
p p2 2
, for
which the function
f xx
xx( ) cos log (cos )= + æèç
öø÷ +é
ëêùûú
 42
12 1
is defined, is
(a) [ , ]0 p (b) æèç
öø÷
p p2 2
,
(c) éëê
öø÷
p p4 2
, (d) 02
,pé
ëêöø÷
86. A body weighing 13 kg is suspended by twostrings 5 m and 12 m long, their other endsbeing fastened to the extremities of a rod 13m long. If the rod be so held that the bodyhangs immediately below the middle point.The tensions in the strings are
(a) 12 kg and 13 kg (b) 5 kg and 5 kg(c) 5 kg and 12 kg (d) 5 kg and 13 kg
87. A pair of fair dice is thrownindependently three times. Theprobability of getting a score of exactly9 twice is
(a) 1/729 (b) 8/9 (c) 8/729 (d) 8/243
88. Consider a family of circles which arepassing through the point (–1, 1) andare touching Xaxis. If ( , )h k are thecoordinates of the centre of the circles,then the set of values of k is given by theinterval
(a) 0 1 2< <k / (b) k ³ 1 2/(c)  £ £1 2 1 2/ /k (d) k £ 1 2/
89. Let L be the line of intersection of theplanes 2 3 1x y z+ + = andx y z+ + =3 2 2. If L makes an angle awith the positive xaxis, then cos aequals
(a) 1 3/ (b) 1 2/ (c) 1 (d) 1 2/
90. The differential equation of all circlespassing through the origin and havingtheir centres on the Xaxis is
(a) x y xydydx
2 2= + (b) x y xydydx
2 2 3= +
(c) y x xydydx
2 2 2= + (d) y x xydydx
2 2 2= 
91. If p and q are positive real numberssuch that p q2 2 1+ = , then the maximum
value of ( )p q+ is
(a) 2 (b)12
(c)1
2(d) 2
92. A tower stands at the centre of a circularpark. A and B are two points on theboundary of the park such that AB a( )=subtends an angle of 60° at the foot ofthe tower and the angle of elevation ofthe top of the tower from A or B is 30°.The height of the tower is
(a) 2 3a/ (b) 2 3a (c) a/ 3 (d) a 3
93. The sum of the series20
020
120
220
320
10C C C C C +  + +K is
(a)  2010C (b)
12
2010C (c) 0 (d) 20
10C
AIEEE Solved Paper 2007 183
94. The normal to a curve at P x y( , ) meets thexaxis at G. If the distance of G from theorigin is twice the abscissa of P, then thecurve is a
(a) ellipse (b) parabola(c) circle (d) hyperbola
95. If  z + £4 3, then the maximum value of
 z + 1 is
(a) 4 (b) 10 (c) 6 (d) 0
96. The resultant of two forces P N and 3 N isa force of 7 N. If the direction of the 3 Nforce were reversed, the resultant wouldbe 19 N. The value of P is
(a) 5 N (b) 6 N (c) 3 N (d) 4 N
97. Two aeroplanes I and II bomb a target insuccession. The probabilities of I and IIscoring a hit correctly are 0.3 and 0.2,respectively. The second plane will bombonly if the first misses the target. Theprobability that the target is hit by thesecond plane is
(a) 0.06 (b) 0.14 (c) 0.2 (d) 0.7
98. If D x
y
= ++
½
½
½½
½
½
½½
1
1
1
1
1
1
1
1
1
for x y¹ ¹0 0, , then
D is
(a) divisible by neither x nor y
(b) divisible by both x and y
(c) divisible by x but not y
(d) divisible by y but not x
99. For the hyperbolax y2
2
2
21
cos sina a = ,
which of the following remains constantwhen a varies?
(a) Eccentricity(b) Directrix(c) Abscissae of vertices(d) Abscissae of foci
100. If a line makes an angle ofp4
with the
positive directions of each of Xaxis andYaxis, then the angle that the line makeswith the positive direction of the Zaxis is
(a) p /6 (b) p /3 (c) p /4 (d) p /2
101. A value of C for which the conclusion ofMean Value Theorem holds for thefunction f x xe( ) log= on the interval [1, 3]is
(a) 2 3log e (b)12
3loge
(c) log3 e (d) loge 3
102. The function f x x x( ) tan (sin cos )= +1 is
an increasing function in
(a) ( / , / )p p4 2 (b) ( / , / )p p2 4
(c) ( , / )0 2p (d) ( / , / )p p2 2
103. Let A =é
ë
êêê
ù
û
úúú
5
0
0
5
0
5
5
aa
aa .
If A2 25= , then a equals
(a) 52 (b) 1
(c)15
(d) 5
104. The sum of the series12
13
14! ! !
 +  K
upto infinity is
(a) e2 (b) e1
(c) e 1
2 (d) e+ 1
2
105. If u and v are unit vectors and q is theacute angle between them, then 2 3u v´is a unit vector for
(a) exactly two values of q(b) more than two values of q(c) no value of q(d) exactly one value of q
106. A particle just clears a wall of height b at adistance a and strikes the ground at adistance c from the point of projection.The angle of projection is
(a) tan1 bac
(b) 45°
(c) tan( )

1 bca c a
(d) tan1 bca
107. The average marks of boys in a class is 52and that of girls is 42. The average marksof boys and girls combined is 50. Thepercentage of boys in the class is
(a) 40 (b) 20(c) 80 (d) 60
184 JEE Main Solved Papers
108. The equation of a tangent to the parabolay x2 8= is y x= + 2 . The point on this line
from which the other tangent to theparabola is perpendicular to the giventangent is
(a) ( , )1 1 (b) (0, 2)(c) (2, 4) (d) ( , )2 0
109. If (2, 3, 5) is one end of a diameter of thespherex y z x y z2 2 2 6 12 2 20 0+ +    + = ,
then the coordinates of the other end ofthe diameter are
(a) ( , , )4 9 3 (b) (4, –3, 3)(c) (4, 3, 5) (d) (4, 3, –3)
110. Let a i j k b i j k= + + =  +$ $ $ , $ $ $2 and
c i j k= +  x x$ ( )$ $2 . If the vector c lies in
the plane of a and b, then x equals
(a) 0 (b) 1 (c) – 4 (d) – 2
111. Let A h k( , ), B( , )1 1 and C( , )2 1 be the vertices
of a right angled triangle with AC as itshypotenuse. If the area of the triangle is 1,then the set of values which ‘k’ can take isgiven by
(a) , 1 3 (b) 0, 2(c) –1, 3 (d) –3, –2
112. Let P Q=  =( , ), ( , )1 0 0 0 and R = ( , )3 3 3 be
three points. The equation of the bisectorof ÐPQR is
(a) 3 0x y+ = (b) x y+ =32
0
(c)3
20x y+ = (d) x y+ =3 0
113. If one of the lines ofmy m xy mx2 2 21 0+   =( ) is a bisector of
the angle between the lines xy = 0, then m
is
(a)  12
(b) 2 (c) ± 1 (d) 2
114. Let F x f x fx
( ) ( )= + æèç
öø÷
1, where
f xt
tdt
x( )
log.=
+ò1 1Then, F e( ) equals
(a)12
(b) 0 (c) 1 (d) 2
115. Let f R R: ® be a function defined by
f x x x( ) min ,  = + +1 1 . Then, which ofthe following is true ?
(a) f x( ) ³ 1for all x RÎ(b) f x( ) is not differentiable at x = 1(c) f x( ) is differentiable everywhere(d) f x( ) is not differentiable at x = 0
116. The function f R R: / 0 ® given by
f xx e x
( ) = 
1 2
12
can be made continuous at x = 0 bydefining f( )0 as
(a) 2 (b) –1(c) 0 (d) 1
117. The solution for x of the equation
2 2 12
x dt
t tò
= p / is
(a) 2 (b) p(c) 3 2/ (d) 2 2
118. ò +dx
x xcos sin3equals
(a)12 2 12
log tanx
C+æèç
öø÷ +p
(b)12 2 12
log tanx
Cæèç
öø÷ +p
(c) log tanx
C2 12
+æèç
öø÷ +p
(d) log tanx
C2 12
æèç
öø÷ +p
119. The area enclosed between the curvesy x2 = and y x= is
(a)23
(b) 1
(c)16
(d)13
120. If the difference between the roots of theequation x ax2 1 0+ + = is less than 5,
then the set of possible values of a is
(a) ( , )3 3 (b) ( , ) ¥3
(c) ( , )3 ¥ (d) ( , )¥  3
AIEEE Solved Paper 2007 185
Solutions
Physics
1. / To determine the position, velocity etc; at first,we write the general representation of waveand then compare the given wave equationwith the general wave equation.
Given, x t= ´ ( )cos2 10 2 p
This gives, a = ´ 2 10 2m = 2 cm
At t = 0,
x = 2 cm
i.e., the object is at positive extreme, so to acquiremaximum speed (i.e., to reach mean position) it
takes14
th of time period.
\ Required time = T4
where, w p p= =2T
Þ T = 2 s
So, required time = =T4
24
= 0 5. s
2. For given circuit current is lagging the voltage byp/2. So, circuit is purely inductive and there is nopower consumption in the circuit. The work doneby battery is stored as magnetic energy in theinductor.
3. / The potential at any point due to system ofcharges is the sum of potentials due toindividual charges, that does not be alteredwhile the changes as some of charges areinterchanged. Potential at A due to charge at O.
VOAA =
4 e1 10
0
3
p( )
=e
×+
14
10
2 20
3
2 2p( )
( ) ( )
=1
410
20
3
pe( )
Applying distance formula to determine length ofOA.
Answers
1. (a) 2. (b) 3. (b) 4. (d) 5. (d) 6. (c) 7. (c) 8. (b) 9. (b) 10. (d)
11. (d) 12. (b) 13. (b) 14. (d) 15. (d) 16. (b) 17. (b) 18. (d) 19. (a) 20. (d)
21. (c) 22. (c) 23. (b) 24. (c) 25. (d) 26. (a) 27. (a) 28. (b) 29. (c) 30. (d)
31. (a) 32. (a) 33. (a) 34. (c) 35. (a) 36. (c) 37. (d) 38. (d) 39. (a) 40. (a)
41. (d) 42. (c) 43. (c) 44. (a) 45. (d) 46. (d) 47. (a) 48. (a) 49. (a) 50. (d)
51. (c) 52. (d) 53. (b) 54. (d) 55. (c) 56. (d) 57. (c) 58. (b) 59. (d) 60. (a)
61. (b) 62. (c) 63. (b) 64. (b) 65. (d) 66. (d) 67. (a) 68. (a) 69. (a) 70. (c)
71. (c) 72. (d) 73. (a) 74. (c) 75. (d) 76. (c) 77. (c) 78. (c) 79. (a) 80. (d)
81. (d) 82. (b) 83. (d) 84. (c) 85. (d) 86. (c) 87. (d) 88. (b) 89. (a) 90. (c)
91. (d) 92. (c) 93. (b) 94. (d) 95. (c) 96. (a) 97. (*) 98. (b) 99. (d) 100. (d)
101. (a) 102. (b) 103. (c) 104. (b) 105. (d) 106. (c) 107. (c) 108. (d) 109. (a) 110. (d)
111. (c) 112. (a) 113. (c) 114. (a) 115. (c) 116. (d) 117. (*) 118. (a) 119. (c) 120. (a)
NOTE (*) None of the given option is correct.
Y
B
A
X(0,0) (2,0)
10–3
CµO
2, 2 ))
X ′
Y ′
Potential at B due to charge at O
VOBB =
e×
14
10
0
3
p( )
= ×1
410
20
3
pe( )
Distance formula is used to determine (OB).
So, V VA B = 0
4. Ratio of energy stored in the capacitor and thework done by the battery
= =12 1
2
qV
qV
5. The circuit is shown in figure below
Rise of current in LR circuit is given by
I I e t=  0 1( )/ t
where IER0 = = =5
51A
Now, t = = =LR
105
2 s
After 2s, i.e., at t = 2 s
Rise of current, I e=  ( )1 1 A (Q =  = t / t 22
1)
6. Current density Ji
a=
p 2
From Ampere’s circuital law
B d i× = ×ò l m0 enclosed
For r a<B r J r´ = ´ ´2 0
2p m p
Þ Bi
a
r= ´mp
02 2
At r a= /2, Bi
a10
4= m
p
For r a> , B r i Bi
r´ = Þ =2
200p m mp
At r a= 2 , Bi
a20
4= m
p
So,B
B1
2
1=
7. Using Ampere’s circuital law the magnetic field at
any point inside the pipe is zero. (QInet = 0)
8. Binding energy
BE = DM c2
= ( )M Mnucleus nucleons c2
=  ( )M M M cp no 8 9 2
9. In gamma ray emission the energy is releasedfrom nucleus, so that nucleus get stabilised.
10. For Vi < 0 the diode is reverse biased and hence,offer infinite resistance, so circuit would be like asshown in Fig. (2) given below and Vo = 0.
For Vi > 0, the diode is forward biased and circuitwould be as shown in Fig. (3) given below andV Vo i= .
Hence, the option (d) is correct.
11. According to deBroglie equation, the momentumof the photon,
ph h
c= =
ln
where, h=Planck’s constant
12. The velocity is given as
v v gt ft= + +02
i.e.,dxdt
v gt ft= + +02
Þ dx v gt ft dt= + +( )02
On integrating0 0
1
02x
dx v gt ft dtò ò= + +( )
= + +é
ëêù
ûúv t gt f
t0
23
0
112 3
Þ x v g f= + +0 2 3/ /
AIEEE Solved Paper 2007 187
10H 5Ω
5V
r
a
Vi Vo
Fig. (2)
Vi Vo
Fig. (3)
5V
–5V
RLViVo
Fig. (1)
13. Let the each side of square lamina is d.
So, I IEF GH= (due to symmetry)
and I IAC BD= (due to symmetry)
Now, according to theorem of perpendicular axes,
I I IAC BD+ = 0
Þ 2 0I IAC = …(i)
and I I IEF GH+ = 0
Þ 2 0I IEF = …(ii)
From Eqs. (i) and (ii), we get I IAC EF=
\ I Imd
AD EF= +2
4(using theorem of parallel axis)
= +md md2 2
12 4as I
mdEF =æ
èç
öø÷
2
12
So, Imd
IAD EF= =2
34
14. Given oscillation is x x t= 0 4cos ( / )w p
Acceleration, ad x
dt=
2
2=  w w p2
0 4x tcos ( / )
= +w w p20 3 4x tcos ( / )
So, amplitude A x= w20
and phase angle d p= 3 4/
15. Direction ofE reverses asE is vector quantitiy whilemagnitude remains same and V remainsunchanged because the net potential at the centreis the sum of potential of four changes.
16. The halflife for element X
T X Y1 2/ ( ) ( )= twhere, t ( )Y is mean life of element Y.
Þ 0 693 1.l lX Y
= Þ l lY
X=0 693.
Þ l lY X>So, Y will decay faster than X.
(Ql is decay constant)
17. For Carnot engine using as refrigerator
W QT
T= æ
èçöø÷2
1
2
1
It is given, h = 110
Þ h = 1 2
1
T
T
Þ T
T2
1
910
= = =+
Q
Q
Q
Q2
1
2
2 10
So, Q2 90= J (as, W = 10 J)
18. The number of free electrons for conduction issignificant only in Si and Ge, but small in C, as C isan impurity.
19. As v of charged particle is remaining constant, itmeans, force acting on charged particle is zero.
So, q q( )v B E´ =Þ v B E´ =
Þ vE B= ´
B2
20. As the electric field is given by
E i j k=  ¶¶
 ¶¶
 ¶¶
Vx
Vy
Vz
$ $ $
Along xdirection, EVxx =  ¶
¶= 
é
ëê
ù
ûú
ddx x
20
42
=
40
42 2
x
x( )
Þ E x at x = =4109
mm V/ mm
and is along positive Xdirection.
21. Emission spectrum would be found when electronmakes a transition from higher energy level tolower energy level.
Frequency of emitted photon is proportional tochange in energy of two energy levels, i.e.,
n = æ
èç
ö
ø÷RcZ
n n
2
12
22
1 1
So, emitted energy will be maximum for maximum
v and hence, emitted photon has maximum
frequency when there is a transition of electron
from n=2 to n=1. While in the transition, from n=1
to n=2, energy will be absorbed.
22. Acceleration of system, aF
m M=
+
So, force acting on mass m ma= =+
mFm M
23. Power of a lens is reciprocal of its focal length.
Power of combined lens is
P P P= +1 2 =  + = 15 5 10D
\ fP
= =
1 10010
cm
f =  10 cm
188 JEE Main Solved Papers
m Mk
F
D
G
A E B
H
CF
d
O
24. Let temperature at the interface be T.
For part AB,
Rate of heat transmission,Q
t
T T K A1 1 1
1
=( )
l
For part BC,
Rate of heat transmission,Q
t
T T K A2 2 2
2
= ( )
l
Here, A is area of crosssections.
At equilibrium,Q
t
Q
t1 2=
\ ( ) ( )T T K A T T K A1 1
1
2 2
2
 = l l
Þ TT K T K
K K= +
+1 1 2 2 2 1
1 2 2 1
l l
l l
25. Let intensity of sound be I and I¢.Loudness of sound initially
b10
10= æèç
öø÷
logII
Later, b20
10= ¢æèç
öø÷
logII
Given, b b2 1 20 =
10 10 200 0
log logII
II
¢æèç
öø÷
 æèç
öø÷
=
Þ 10 200logII
I
I¢æ
èçöø÷ ´ æ
èçöø÷
éëê
ùûú
=
Þ 10 20logII¢æ
èçöø÷ = Þ log
II¢æ
èçöø÷ = 2
Þ II¢ = 102 Þ I I¢= 100
26. According to Mayer’s relation,
C CRm
Rp V = =
28(QC C Rp V = for 1 mole)
As molecular weight of N2 gas is 28.
27. In case of motion of a charged particleperpendicular to the motion, i.e., displacement,the work done
W Fds= × = =ò òF ds cos q 0 (as q = °90 )
and by workenergy theorem, W = DKE, the kineticenergy and hence, speed v remains constant. Butv changes because of change in its direction, somomentum changes.
28. The magnetic fieldinduction at a point P, at adistance d from O in adirection perpendicular tothe plane ABCD due tocurrents through AOB andCOD are perpendicular toeach other, is
Hence, B B B= +12
22
= æèç
öø÷
+ æèç
öø÷
é
ëê
ù
ûú
mp
mp
0 12
0 22 1 2
4
2
4
2I
d
I
d
/
= +mp
012
22
2 dI I( )
29. From R R tt = +0 1( )a
5 1 500= +R ( )a …(i)
and 6 1 1000= +R ( )a …(ii)
\ 56
1 501 100
= ++
aa
Þ a = 1200
Putting value of a in Eq (i), we get
5 1 50 1 2000= + ´R ( / )
\ R0 = 4 W
30. On introduction and removal and again onintroduction, the capacity and potential remainsame. So, net work done by the system in thisprocess
W U Uf i=  = 12
12
2 2C V C V = 0
31. According to Millikan’s oil drop experiment,
electronic charge is given by, qr v v
E= +6 1 2ph ( )
(Qviscous force = electric force Þ 6p hsv qE= ')
which is independent of g.
So,electronic charge on the moonelectronic charge on the earth
= 1
32. In this question distance ofcentre of mass of new discfrom the centre of mass ofremaining disc is aR.
Mass of remaining disc
=  =M M M/ /4 3 4
\  + =34 4
0M
RM
Ra
Þ a = 1 3/
NOTE The given distance must be aR for real approach tothe solution.
AIEEE Solved Paper 2007 189
C
AP
D
B
d
I1
I2O
T1l1 l2 T2
K1 K2A B C
T
OO1O2
aR
R
33. / While rolling a body on a incline, has threetypes of energy. Kinetic energy, rotationalkinetic energy and potential energy when thereis no external force acting on the system and allthe external forces are conservative, the totalmechanical energy of the system will remainconserved. Assuming that no energy is used upagainst friction, the loss in potential energy isequal to the total gain in the kinetic energy.
Thus, from workenergy theorem,
Mgh I v R Mv= +12
12
2 2 2( / ) Qw =æèç
öø÷
vR
or12
2 2v M I R Mgh( / )+ =
or vMgh
M I R
gh
I MR
2
2 2
2 2
1=
+=
+/ /
If s is the distance covered along the plane, then
h s= sin q
vgs
I MR
2
2
2
1=
+sin
/
q
Now, v as2 2=
\ 22
1 2as
gs
I MR=
+sin
/
qor a
g=+
sinq1 I / MR2
34. According to the principle of conservation ofangular momentum, in the absence of externaltorque, the total angular momentum of the systemis constant.
35. / Consider all the force belonging to the systemand work done by all the forces with propersign. Then, apply workenergy theorem.
Given, k =10,000 N/m
Force of kinetic friction, fk = 15 N
Mass of block = 2 kg
Initial velocity of the block = 4 m/s
Suppose block gets the state of rest aftercompressing the spring.
Applying workenergy theorem,
i.e., work done by all the forces
= change in kinetic energy
f x kx mvk ( )+ = 12
12
02 2
Þ 1512
1000012
2 42 2x x+ ´ ´ = ´ ´
Þ 15 5000 162x x+ =
Þ 5000 15 16 02x x+  =
Thus, is quadratic in terms of x, which on solvinggives the value of x = 9 34. cm
As x =  ±  ´ ´ ´
15225 4 5000 16
2 5000
=  ± +15
225 32000010000
=  ±1532022510000
=  ±15 32 0225.
=  ±15 5 66. =  +15 5 66.
= 9 34. cm
36. / The horizontal component of velocity inprojectile motion is always constant, sincethere is no acceleration in this direction.
The KE of projectile of point of projection is
K mv= 12
2
Kinetic energy at highest point
( )KE H mv= 12
2 2cos q
= Kcos2 q = °K(cos )60 2
= K4
37. As phase difference = ´2pl
path difference
i e. ., f = ´26
pl
l = p3
As I I= fmax cos ( / )2 2
orI
Imax
= fcos ( / )2 2
orII0
2 6= cos ( / )p (where, I Iomax = )
= 34
38. The frequencies of oscillation in this situations isgiven by
Q f f f= +1 2 (same direction)
=   =  +k x k x k k x1 2 1 2( )
Þ f k x k k x¢=  ¢ = +( )1 2
Þ TMk
=¢
2p Þ fkm
= ¢12p
Þ fk k
m= +1
21 2
p
and fk k
mf¢ = × + =1
22 21 2
p(Qk k k k1 1 2 24 4= =, )
190 JEE Main Solved Papers
h
θ
C
s
C
3
v
m
k1
k2
39. From first law of thermodynamics,
Q U W= +DFor path iaf, 50 20= +DU
\ DU U Uf i=  = 30 cal
For path ibf, Q U W= +Dor W Q U=  D = 36 30 = 6 cal
(QDU is same being path independent function)
40. The average kinetic energy of particle
= 14
2 2ma w
= 14
22 2ma ( )pn (Q w p pn= =22
T)
= p n2 2 2ma
Chemistry
41. A B AB2 2+ a2
Ea( )forward kJ mol= 180 –1
Ea( )backward = 200 kJ mol 1
In the presence of catalyst,
Ea( )forward =  =180 100 80 kJ mol 1
Ea( )backward =  =200 100 100 kJ mol 1
DH Ea= ( )forward  Ea( )backward
= 80 100 =  20 kJ mol 1
42. Cell is completely discharged, it meansequilibrium gets established, Ecell = 0
Zn  Zn (1M) Cu (1M)Cu2+ 2+
Cell reaction : Zn + Cu Zn + Cu2+ 2+a
Keq
2+
2+
[Zn ]
[Cu ]=
We know, E En
Kcell cell eq= °  0 0591.log
En
K° =cell eq0 0591.
log
or 1.10 eq= 0 05912
.log K
or log10 Keq1.10 20.0591
= ´
log10[Zn ]
[Cu ]
2.200.0591
2+
+ =
or log10[Zn ]
Cu
2+
+ = 37.23
or[Zn ]
[Cu ]
2+
+ = 1037 23.
43. Aqueous buffered solution of HA
50% H A is ionised Þ[ ] [H A A=  ]
Buffer solution of weak acid HA ® acidic buffer
pH = p + log[ ][H ]
–
KA
Aa
or pH p 4.5= =Ka
pOH p – pH= Kw
pOH =  =14 4 5 9 5. .
Alternate Solution
For buffer solution
pH log[Salt][Acid]
= +PKa = +4.5 log[Salt][Acid]
as [HA] is 50% ionised so [Salt]=[Acid]
hence, pH = + =4 5 0 4 5. .
Since, pH p 14+ =OH ,
therefore, pOH 14 4.5 9.5=  =
44. 2 A B+ ¾® products
[B] is doubled, halflife didn’t change
Halflife is independent of change in concentrationof reactant Þ first order
First order w.r.t. B
[A] is doubled, rate increased by two times
Þ First order w.r.t. A
Hence, net order of reaction = + =1 1 2
Unit for the rate constant = conc.( )1 1 n t
= × × (mol L ) s–1 –1 1 = ×L mol s–1 –1
45. 4f and 5f belongs to different shell, thereforeexperience different amount of shielding.
46. Cl is a weak ligand but Cl cause the pairing ofelectron with large Pt2+ and consequently givedsp2 hybridisation and square planar geometry.
Other three are sp3 hybridised having tetrahedral
shape.
47. / A molecule is optically active if it containschiral centre and does not have plane ofsymmetry. So, check chiral centre and plane ofsymmetry in molecules. The molecule, which isoptically active, has chiral centre, is expectedto rotate the plane of polarised light.
AIEEE Solved Paper 2007 191
HO
CHO
* H
CH OH2
One chiral centre Þ optically active
Two chiral centres, but plane of symmetry withinmolecule Þ optically inactive
48. Primary structure involves sequence of aaminoacids polypeptide chain. Secondary structureinvolves ahelical backbone bpleated sheet likestructures. These structures are formed as a resultof Hbonding between different peptide groups.
49. CH C CH + HBr3 ¾ ºº ¾® ¾½
CH C
Br
==CH3 2
¾ ¾¾¾¾¾Rearrangement
H C C ==CH3 2¾½
+H
Br
¬¾¾¾+ HBr
¾® ¾ ¾½
CH C
Br
CH3
+
3
more stablecarbocation
¾®Br H C C
Br
Br
H3 3
2,2 dibromopropane
¾ ¾½
½

C
CH CH==CHBr3 ¾ ¾® ¾ ¾HBr3 2 2CH CH CHBr
CH CH CH CHBr2HBr3 2ºº ¾¾®
CH CH == CH CH CHBr CH3 2HBr
3 3¾ ¾¾® ¾ ¾
50. CH CH NH + CHCl + KOH 3H O3 3 2 3 2® + +A B
CHCl + KOH CCl +KCl+ H O3 2[ ]
2¾® ··
B
CH CH NH + CCl CH CH NC3 2 2 22KOH
3 2
[
arbylamine·· ¾ ®¾¾
cA ]
+ 2KCl +2H O2
51.
52.
¾NO2 group withdraw electron from the ring,shows M effect makes ring electron deficient,thus, deactivates ring for electrophilic substitution.
53. / Bond order and magnetic behaviour depends onthe electronic distribution in molecularorbitals. So, write the electronic configurationof molecules and calculate the bond order andmagnetic behaviour. NO NO+¾®
(NO) Total e = 15
s s s s1 1 2 22 2 2 2s s s s* * , s p p2 2 22 1 1 1 1p p pz x y+ +=
p p*2 21p px y=
Paramagnetic
Bond order =  =10 52
2 5.
(NO)+
Total e = 14
s s s s p1 1 2 2 22 2 2 2 1 1s s s s px* * + = +2 21 1 2p py zs
Diamagnetic
Bond order =  =10 42
3
Electron is taken away from nonbondingmolecular orbital, that’s why bond order increasesand nature changes from paramagnetic todiamagnetic.
54. The actinoid (5felements) exhibits more numberof oxidation states in general than the lanthanoidbecause 5f orbitals extend farther from the nucleusthan the 4f orbitals.
55. / Partial pressure depends on the mole fraction.So, find the mole fraction, then calculatepartial pressure.
Suppose the equal mass of methane and oxygen
= =w 1 g
Mole fraction of oxygen =+
w
w w
// /
3232 16
= =1
323 32
13/
Let the total pressure = p
Pressure exerted by oxygen (partial pressure)
= ´X pO total2= ´P
13
192 JEE Main Solved Papers
H N2
H
Ph Ph
H
NH2
Cl ,FeCl2 3
(Cl )+
CH3
Cl
CH3
Cl
CH3
toluene pchloro toluene
ochloro toluene
+
chlorination,electrophilicsubstitution
N+
O
O–
56. Solution is isotonic.
Þ C RT C RT1 2=C C1 2=
Density of both the solutions are assumed to be equalto 1.0 g cm 3 .
Þ Molality = Molarity = 5.25%
QIn 100 g, 5.25 g of substance is present.
\ In 1000 g, 52.5 g of substance is found.
Hence,52 5 15
60.
M= ,
M = molecular mass of the substance
M = ´52 5 6015
. = 210 g mol1
Alternate Solution 5.25% solution means 5.25 g ofsubstance dissolved in 100 g of solution. 1.50%solution means 1.50 g of urea dissolved in 100 g ofsolution.
Q Solution is isotonic with solution of urea
\ p psub urea=C RT C RTsub urea=
Thus, C Csub urea=M
M V
M
M Vsub
sub sub
urea
urea urea´=
´5.25
1001.50
60 100subM ´=
´
Msub 210= g mol1
57. H O H O2 2( ) ( )l g¾®
Dng =  =1 0 1
D D DE H n RTg=  =  ´ ´ ´ 41 1 373 10 38.3
( . )R = ´ 8 3 10 3
= 37 9. kJ mol 1
58. / K sp depends on the solubility or vice versa.
Solubility is the maximum mass of solutes
( )in grams dissolved in one litre of solubility.Therefore, find the solubility from K sp and then,calculate mass.
AgIO ( ) Ag ( ) IO ( )3 3s aq aqa+ +
Let solubility of AgIO3 be S
Ksp+
3–[Ag ][IO ]=
10 10 8 2. ´ = S
or S = ´ 1 10 4 mol/L
In 1000 mL mol of AgIO3 dissolved = ´ 1 10 4 mol
In 100 mL of mole of AgIO3 dissolved = ´ 1 10 5 mol
Mass of AgIO3 in 100 mL = ´ ´1 10 2835
= ´ 2 83 10 3. g
59. Activity µ N
NN
n
0
12
= æèç
öø÷
or1
1012
= æèç
öø÷
n
or 10 2= n
Taking log on both sides
log log10 2= n
n = =10 301
3 32.
.
Time = ´n halflife
= ´3 32 30.
= 99 6. days
» 100 days
60. Chiral conformation will not have plane ofsymmetry. Twisted boat is chiral as it does nothave plane of symmetry.
61. S 2N reactions are greatly controlled by steric
factor.
R X¾ ¾°
CH21
> R X22
CH¾°
> R X33
C ¾°
S 2N reactivity decreases as bulkyness of alkylgroup increases.
62.
63. n l+ = 5 maximum, thus has the highest
energies.
64. Hydrogen bond is strongest in HF due tohigher electronegativity of F.
65. 2Al 6HCl 2Al Cl1 mol
3( ) ( ) ( ) ( )s aq aq aq+ ¾® ++ 6
+ 3H2 ( )g
1 mole of HCl consumed for the production of0.5 mole of H2 gas
At STP, 1 mol = 22.4 L
0.5 mol = 22 4 0 5. .´ = 11 2. L
AIEEE Solved Paper 2007 193
CH CH OH3 2 CH CH2 2
CH__
CH__
CH3 2 2
CH CH CH + Mg(OH)I3 2 2
OMgI
OH
O
CH CH MgI3 2
__I
P + I2
‘ ’A
‘ ’B‘ ’C
‘ ’Dnpropyl alcohol
H O2
Mg, ether
H__
C__
H
66. Ammonium sulphate produce H+ ions onhydrolysis. This will increase the acidity.
(NH ) SO 2NH +SO4 2 4
+
4cidic
42–¾®
A
½¯
½H O2
NH OH H4 + r
67. In an isolated system where either mass andenergy are not exchanged with surrounding for thespontaneous process, the change in entropy ispositive.
68. ZA
Z
AX X n¾® + 10
1
Isotopes are species having same number ofproton, but different number of neutron.
69. We know from Kohlrausch’s law,
l° = + l°  l°CH COOH CH COON HCl NaCl3 3l° a
70. In aqueous solution, basicity order
Dimethylamine >methylamine2 1° °
> trimethylamine > aniline1°
71. Any aliphatic carbon with hydrogen attached to it,
in combination with benzene ring, will be oxidised
to benzoic acid by KMnO / H4+.
72.
73. The correct option is O22. This species has 18
electrons, which are filled in such a way that allmolecular orbitals are fully filled, so diamagnetic.
s s s s1 1 2 22 2 2 2s s s s* *, s2 2pz , p p2 22 2p px y= ,
p p* *2 22 2p px y=
74. Due to inert pair effect, the stability of +2 oxidationstate increases as we move down this group.
\ Si <Ge < Sn < Pb2 2 2 2X X X X
75. Br2 reacts with hot and strong NaOH to give NaBr,
NaBrO3 and H O2 .
76. Higher the charge/size ratio, more is the polarisingpower.
\ K < Ca < Mg < Be+ 2+ 2+ 2+
77. Molarity=10 density weight of solutemolecular weight
´ ´of the solute
Density = ´´
= 3 60 9810 29.
1.21gmL 1
Alternate Solution
Let the density of solution be ’d‘
Molarity of solution given = 3 6.
i.e., 1 litre of solution contains 3.6 moles of H SO2 4
or 1 litre of solution contains 3.6 ´ 98 g of H SO2 4.
Since, the solution is 29% by mass.
Density =´
molecular wt. of the soluteweight of solute10
= ´´
3 6 9810 29. = 1.21 gmL1
78. K k k= ´1 2 = ´ ´ ´ 10 10 5 0 105 10. .
= ´ 5 10 15
79. According to Raoult’s law
p p X p XA A B B= ° + °290 200 0 4 0 6= ´ + ° ´. .P B
p B° = 350
80. D DS
HT
=
DS = 160.2 J/ K
DH = ´1791 103. J / mol
T = ´1791 103./J / mol
160.2 J M= »1117 9 1118. K K.
Mathematics
81. Since, each term is equal to the sum of next twoterms.
\ ar ar arn n n += +1 1
Þ 1 2= +r r Þ r r2 1 0+  =
Þ r = 5 12
Q r ¹  æèç
öø÷
5 12
82. Q sin æèç
öø÷ + æ
èçöø÷ =1 1
554 2
xcosec
p
Þ sin sin æèç
öø÷ + æ
èçöø÷ =1 1
545 2
x p
Þ sin sin æèç
öø÷ =  æ
èçöø÷
1 1
5 245
x p
Þ sin cos æèç
öø÷ = æ
èçöø÷
1 1
545
x
Þ sin sin æèç
öø÷ = æ
èçöø÷
1 1
535
x
Þ x = 3
194 JEE Main Solved Papers
7
6
54
3
2
1
3ethyl4, 4dimethylheptane
83. / Use the general term in the expansion of( )x y n+ is T C x yr
nr
n r r+
=1 .
Since, in binomial expansion of ( )a b n , n ³ 5, thesum of fifth and sixth terms is equal to zero.
\ n n n nC a b C a b44 4
55 5 0  +  =( ) ( )
Þ nn
a bn
na bn n!
( )! !!
( )! !× 
= 
4 4 5 504 4 5 5
Þ nn
a ba
nbn!
( )! !×
ìíî
üýþ=
5 4 4 505 4
Þ ab
n=  45
84. Required number of ways = ´ ´124
84
44C C C
=´
´´
´128 4
84 4
1!
! !!
! != 12
4 3
!
( !)
85. 42 x is defined for æ
èçöø÷
p p2 2
, .
cos æèç
öø÷
1
21
xis defined, if  £  £1
21 1
x.
Þ 02
2£ £x Þ 0 4£ £x
and log (cos )x is defined, if cos x > 0.
Þ  < <p p2 2
x
Hence, f xx
xx( ) cos log (cos )= + æèç
öø÷ + 4
21
2 1
is defined, if x Îéëê
öø÷0
2, .
p
86. / If three coplanar forces are in equilibrium, theneach force is proportional to the sine of theangle between the other two forces.
Q OC CA CB= = Þ Ð = ÐAOC OAC
and Ð = ÐCOB OBC
\ sin sinq = =A5
13and cos q = 12
13
Now, by Lami’s theorem,T T1 2
180 90sin ( ) sin ( )° =
° +q qÞ T T1 2
sin cosq q=
Þ T T1 2cos sinq q=
Þ T T1 21213
513
æèç
öø÷ = æ
èçöø÷ Þ T T1 2
512
= æèç
öø÷
Also, T T1 2 13sin cosq q+ =
Þ T25
125
131213
13× +æèç
öø÷ =
Þ T2169
12 1313
×æèç
öø÷ =
Þ T2 12= kg and T1 5= kg
87. Probability of getting score 9 in a single throw
= =436
19
Probability of getting score 9 exactly in double
throw = ´ æèç
öø÷ ´ =3
2
219
89
8243
C
88. Equation of circle which touches xaxis andcoordinates of centre are ( , ),h k is
( ) ( )x h y k k +  =2 2 2
QIt is passing through (– 1, 1), then
( ) ( )  +  =1 12 2 2h k k
Þ h h k2 2 2 2 0+  + =Q For real circle, D ³ 0 Þ 2 1 0k  ³
Þ k ³ 12
89. / Since, the line lies on both the given planes,then the normal to the planes areperpendicular to the line L.
If direction cosines of L are l, m, n, then
2 3 0l m n+ + = K(i)
and l m n+ + =3 2 0 K(ii)
On solving Eqs. (i) and (ii), we get
l m n3 3 3
=
=
∴ l m n: : : := æèç
öø÷
1
3
1
3
1
3
⇒ l = 1
3Þ cos a = 1
3
90. / Use the general equation of circle
x y gx fy c2 2 2 2 0+ + + + = and eliminate g.
General equation of all such circles which passthrough the origin and whose centre lie on xaxis, is
x y gx2 2 2 0+ + = K(i)
On differentiating w.r.t. x, we get
2 2 2 0x ydydx
g+ + =
Þ 2 2 2g x ydydx
=  +æèç
öø÷
AIEEE Solved Paper 2007 195
p  q/2qp/2  q q 5 cm
12 cm
O13 kg
A
C
B
T1sinq T
1cosq
T1
T2
On putting the value of 2 g in Eq. (i), we get
x y x ydydx
x2 2 2 2 0+ +  æèç
öø÷ =
Þ x y x xydydx
2 2 22 2 0+   =
⇒ y x xydydx
2 2 2= +
which is required equation.
91. / Use AMGM inequality in the positive realnumbers p and q and solve it.
Using AM ³ GM
p qpq
2 2
2+ ³
⇒ pq £ 12
(Qp q2 2 1+ = )
Now, ( )p q p q pq+ = + +2 2 2 2
Þ ( )p q pq+ = +2 1 2
Þ ( )p q+ £ +2 1 1
⇒ p q+ £ 2
92. Let h be the height of a
tower
Q Ð = °AOB 60
\ DOAB is equilateral.
∴ OA OB AB a= = =Now, in DOAC,
tan30° = ha
Þ 1
3= h
aÞ h
a=3
93. / Use binomial theorem of expansion for ( )1 + x n
and solve it for x = 1.
We know that,
( ) ...1 20 200
201
2010
10+ = + + + +x C C x C x
...+ 2020
20C x
On putting x =  1in above expansion, we get
0 200
201
209
2010
2011=  +  +  +C C C C C...
...+ 2020C
Þ 0 200
201
209
2010=  +  +C C C C. . .
 + +209
200C C. . .
Þ 0 2 200
201
209
2010=  +  +( ... )C C C C
Þ 2010
200
201
20102C C C C=  + +( . . . )
⇒ 200
201
2010
2010
12
C C C C + + =...
94. Equation of normal is Y ydxdy
X x =  ( ). It meets
the xaxis at G. Therefore, coordinates of
G x ydydx
= +æèç
öø÷, 0
According to question,
x ydydx
x+ = 2 ⇒ ydydx
x= or ydydx
x=  3
Þ ydy xdx=or ydy xdx=  3
Þ y xC
2 2
2 2= +
ory x
C2 2
232
=  +
Þ x y C2 2 2 =  or 3 22 2x y C+ =
95.  z + £4 3 represents the
interior and boundary ofthe circle with centre at( , )4 0 and radius = 3.From the arganddiagram, maximumvalue of z + 1 is 6.
Alternate Solution
       z z z+ = +  £ + +  £1 4 3 4 3 6
Thus, maximum value of z + 1 is 6.
96. 7 3 2 32 2 2= + + ´ ´P Pcosq ... (i)
and ( ) ( ) ( ) cos19 3 2 32 2 2= +  + ´  ´P P q ...(ii)
On adding Eqs. (i) and (ii), we get
68 2 18 52= + Þ =P P
97. / Firstly, find the probability of incorrect target.
Let the events
A = 1st aeroplane hit the target,
B = 2nd aeroplane hit the target and theircorresponding probabilities are
P A( ) .= 0 3 and P B( ) .= 0 2
P A( ) .= 0 7 and P B( ) .= 0 8
The desired probability
= + +P A P B P A P B P A P B( ) ( ) ( ) ( ) ( ) ( ) ...
= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2)
+ (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + ...
= + + +0 14 1 0 56 0 56 2. [ ( . ) ( . ) ...]
=
éëê
ùûú
0 141
1 0 56.
.= = =014
0 44722
0 32..
.
From above, it is clear that no option is correct.
196 JEE Main Solved Papers
a
aa
B
A
C
O30°
60°
h
( 7, 0)( 4, 0)
( 1, 0)
y
x
P
7
3O
q3
19
98. D xy
= ++
1 1 11 1 11 1 1
On applying C C C C C C2 2 1 3 3 1®  ® and
= =1 0 01 01 0
xy
xy
Hence, D is divisible by both x and y.
99. / Use the eccentricity formula i.e., eb
a= +1
2
2.
The given equation of hyperbola isx y2
2
2
21
cos sin.
a a =
Here, a2 2= cos a and b2 2= sin aCoordinates of foci are (± ae, 0).
Q eb
a= +1
2
2
Þ e = +12
2
sin
cos
aa
= +1 2tan a
Þ e = sec a
Hence, abscissae of foci remains constant when αvaries.
100. Since, a line makes an angle ofp4
with positive
direction of each of Xaxis and Yaxis, therefore
a p b p= =4 4
,
We know that, cos cos cos2 2 2 1a b g+ + =
Þ cos cos cos2 2 2
4 41
p p g+ + =
Þ 12
12
12+ + =cos g
Þ cos2 0g = Þ g = °90
101. Using mean value theorem,
\ ¢ = 
f cf f
( )( ) ( )3 1
3 1⇒ 1 3 1
2c= log log
⇒ c ee
= =23
2 3loglog
102./ Differentiate w.r.t. x and if f x¢ >( ) 0 for given
interval, then the function is increasing.
Q f x( ) tan= 1 (sin cos )x x+
\ ¢ =+ +
f xx x
x x( )(sin cos )
(cos sin )1
1 2
=+æ
èçöø÷
+ +
24
1 2
cos
(sin cos )
x
x x
p
f (x) is increasing, if  < + <p p p2 4 2
x
Þ  < <34 4p p
x
Hence, f (x) is increasing when x Î æèç
öø÷
p p2 4
, .
103./ Firstly, solve A A A2 = × and then A 2 25= .
Q A =é
ëêê
ù
ûúú
5 50 50 0 5
a aa a
\ A25 50 50 0 5
5 50 50 0 5
=é
ëêê
ù
ûúú
é
ëêê
ù
ûúú
a aa a
a aa a
=+ +
+é
ë
êê
ù
û
úú
25 25 5 10 250 5 250 0 25
2 2
2 2a a a a
a a a
 A2
2 2
2 2
25 25 5 10 25
0 5 250 0 25
=+ +
+a a a a
a a a
= +25
25 25 5
0
2
2
a aa
= 625 2a
But  A2 25=
\ 625 252a = Þ  a = 15
104.12
13
14! ! !
 +  K
=  +  + 1 112
13
14! ! !
K = e 1
105. Let (2 3u v´ ) is a unit vector.
 2 3 1u v´ =Þ 6u v     =sin 1q
Þ sin q = 16
(Q   u v= = 1)
Since, q is an acute angle.
Hence, there is exactly one value of θ for which(2 3u v´ ) is a unit vector.
106. a u t= ( cos )a and b u t gt= ( sin )a 12
2
b a ga
u= tan
cosa
a12
2
2 2
Also, cu
g=
2 2sin a
AIEEE Solved Paper 2007 197
u
B
DA a
C
b
c
b aa g
cg=  æ
èçöø÷
tansin
seca a a2
2
22
Þ b aac
=  ´tan tana a2
22
⇒ aac
bæèç
öø÷ =
2
tana
Þ tan( )
a =
bca c a
Þ a =
tan( )
1 bca c a
107./ Total number of boys and girls= Average ´ Number of boys
+ Average ´ Number of girls.
Let the number of boys and girls be x yand .
\ 52 42 50x y x y+ = +( )
Þ 52 42 50 50x y x y+ = +Þ 2 8x y= Þ x y= 4
\ Total number of students in the class
= + = + =x y y y y4 5
\ Required percentage of boys
= ´ =45
100 80yy
% %
108. Since, perpendicular tangent intersect on thedirectrix. So, required point must be on the directrixof the parabola.
Hence, the required point is (– 2, 0).
109. Equation of sphere is
x y z x y z2 2 2 6 12 2 20 0+ +    + =
whose coordinates of centre are (3, 6, 1).
Let the coordinates of the other end of diameter be
(α, β, γ),
thena b g+ = + = + =2
23
32
65
21, ,
Hence, a b= =4 9, and g =  3.
Thus, the coordinates of other point are (4, 9, – 3).
110. Qa i j k= + +$ $ $, b i j k= + +$ $ $2
and c i j k= +  x x$ ( )$ $2 are coplanar.
\x x  
=
2 11 1 11 1 2
0
Þ1 1 2 2 1 1 2 1 2 0 ( ) ( ) ( )     +  + =x x x x
Þ1 2 4 1 2 2 2 0 + + + +  =x x x
Þ 3 2 2 0x x+  + =Þ 2 4x = Þ x = 2
111. QA h k B( , ), ( , )1 1 and C( , )2 1 are the vertices of a right
angled DABC.
Now, AB h k=  + ( ) ( )1 12 2
or BC =  + ( ) ( )2 1 1 12 2
or CA h k=  + ( ) ( )2 12 2
Now, by Pythagoras theorem,
AC AB BC2 2 2= +
4 4 1 22 2+  + + h h k k
= +  + +  +h h k k2 21 2 1 2 1
Þ 5 4 3 2 = h h Þ h = 1 K(i)
Now, given that area of the triangle is 1.
Then, area ( )DABC AB BC= ´ ´12
112
1 1 12 2= ´  +  ´( ) ( )h k
2 1 12 2=  + ( ) ( )h k K(ii)
Putting h = 1from Eq. (i), we get 2 1 2= ( )k
Squaring both sides, we get
4 1 22= + k k or k k2 2 3 0  =
or ( ) ( )k k + =3 1 0
So, k =  1 3,
Thus, the set of values of k is 1, 3.
112. Slope of QR
= 
3 3 03 0
= =3 tanqSlope of the line
QM
= tan23
3p = 
The angle
between PQR is23p
, so line QM makes an angle23p
from positive direction of Xaxis.
Hence, equation of line QM is y x=  3
or 3x y 0+ = .
198 JEE Main Solved Papers
A k k( , )Y
O X
C(2, 1)B(1, 1)
p/32 /3pP
M
Q (0, 0) X
YR (3, 3 3)
( 1,0)
113. Equation of bisectors of lines xy = 0 are y x= ± .
Put y x= ± in my m xy mx2 2 21 0+   =( ) , we get
mx m x mx2 2 2 21 0+   =( )
Þ ( )1 02 2 =m x ⇒ m = ± 1
114. Q f xt
tdt
x( )
log=+ò 11
and F e f e fe
( ) ( )= + æèç
öø÷
1
Þ F et
tdt
t
tdt
e e( )
log log=+
++ò ò1 11 1
Put tt
= 1in second integration.
=+
++ò ò
log log( )
t
tdt
t
t tdt
e e
1 11 1
= =é
ëê
ù
ûúò
log (log )t
tdt
tee
1
2
12
=  =12
112
2 2[(log ) (log ) ]e
115. f x x x( ) min ,  = + +1 1
f x x x R( ) ,= + " Î1
From figure, it is clear that f x( ) ,³ 1 when x RÎ .
116. lim1x
2
10 2x xe®
ìíî
üýþ=  
®lim
( )x
x
x
e x
x e0
2
2
1 2
1
=  +®
lim( )x
x
x x
e
e xe0
2
2 2
2 2
1 2
(using L’ Hospital rule)
=+
=®limx
x
x x
e
e xe0
2
2 2
4
4 41
\ f x( ) is continuous at x = 0, then
lim ( ) ( )x
f x f®
=0
0
Þ 1 (0).= f
117.d
t t
px t
1 22 =ò 2
Þ [sec ] =1
2 2t x p
Þ sec  =1
4 2x
p p
Þ sec =1 34
xp
Þ x =  2
118.dx
x xcos sin+ò 3
=+æ
èçöø÷
ò dx
x x212
32
cos sin
= æèç
öø÷ò1
2 3sec x dx
p
=  +æèç
öø÷ +1
2 2 6 4log tan
xC
p p
= +æèç
öø÷ +1
2 2 12log tan
xC
p
119. Required area, A x x dx= ò ( )0
1
= é
ëêù
ûú23 2
3 22
0
1
xx/ =  =2
312
16
120. Let a and b be the roots of equation
x ax2 1 0+ + = , then
a b+ =  a and ab = 1
Now,   ( )a b a b ab = + 2 4
Þ  a b = a2 4
According to the question,
a2 4 5 < Þ a2 4 5 <Þ a2 9 0 < Þ a Î ( , )3 3
AIEEE Solved Paper 2007 199
y x 1= +y
y x 1=  +
(0, 1)
Ox
Yy  x=
(1, 0)OX
Y ¢
X ¢
y x2=