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Transcript of IIT jee advanced 2013
2013 JEEADVANCED_P1_Code::0 QPaper with & Key &Sol Solutions
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IMPORTANT INSTRUCTIONS: CODE :0A. General:* This booklet is your Question Paper. Do not break the seals of this booklet before being instructed to do
so by the invigilators.* The question paper CODE is printed on the right hand top corner of this sheet and on the back page
(Page No. 44) of this booklet.
* Blank spaces and blank pages are provided in the question paper for your rough work. No additional
sheets will be provided for rough work.* Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic
gadgets are NOT allowed inside the examination hall.
* Write your name and roll number in the space provided on the back cover of this booklet.
* Answers to the questions and personal details are to be filled on a twopart carbonless paper, which isprovided separately. These parts should only be separated at the end of the examination when instructed
by the invigilator. The upper sheet is a machinegradable Objective Response Sheet (ORS) which will be
retained by the invigilator. You will be allowed to take away the bottom sheet at the end of the examination.
* Using a black ball point pen darken the bubbles on the upper original sheet. Apply sufficient pressureso that the impression is created on the bottom duplicate sheet.
* DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.
* On breaking the seals of the booklet check that it contains 44 pages and all the 60 questions and
corresponding answer choices are legible. Read carefully the instruction printed at the beginning ofeach section.
* B. Filling the right part of the ORSThe ORS also has a CODE printed on its left and right parts.Check that the CODE printed on the ORS (on both sheets) is the same as that on this booklet and putyour signature affirming that you have verified this.
* IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET.* Write your Name, Rol No. and the name of centre and sign with pen in the boxes provided on the right
part of ORS. Do not write any of this anywhere else. Darken the appropriate bubble UNDER each digitof your Roll No. in such a way that the impression is created on the bottom sheet.
C. Question Paper FormatThe question paper consisits of three parts ( Physics, Chemistry and Mathematics). Each part consists ofthree sections.
* Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which only ONE is correct
* Section 2 contains 5 Multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE or MORE are correct
* Section 3 Contains 5 questions. The answer to each question is a singledigit integer, ranging from 0 to 9(both inclusive)
D. Marking Scheme* For each question in section 1, you wil be awarded 2 marks if you darken the bubble corresponding to
the correct answer and zero mark if no bubbles are darkened. No Negative marks will be awarded forincorrect answers in this section
* For each Section 2, you will be awarded 4 marks if you darken all the bubble(s). corresponding to onlythe correct answer(s) and zero mark if no bubbles are darkened. in all other cases, minus one (1) markwill be awarded.
* For each question in section 3, you will be awarded 4 marks if you darken the bubble corresponding toonly the correct answer and zero mark if no bubbles are darkened. in all other cases, minus one (1) markwill be awarded.
2013 JEEADVANCED_P1_Code::0 QPaper with & Key &Sol Solutions
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PARTI_PHYSICSSection1
(Only one Option correct Type)This section contains 10 Multiple Choice questions. Each Question has Four choices(A), (B), (C) and (D). Out of Which Only One is correct
1. The work done on a particle of mass m by a force 3/ 2 3/ 22 2 2 2ˆ ˆx yK i j
x y x y
(K being a constant of appropriate dimensions), when the particle is taken from thepoint ,0a to the point 0, a along a circular path of radius a about the origin inthe x y plane is
A) 2Ka B) K
a C) 2
Ka D) 0
Ans : D
Sol : 3krFr
i.e conservative force from the origin at equal distances potentials will be
constant.work done = U 0 Ans : 0
2. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure, One of the blocks hasthermal conductivity and the other 2 . The temperature difference between theends along the xaxis is the same in both the configurations. It takes 9 s to transporta certain amount of heat from the hot end to the cold end in the configuration I. Thetime to transport the same amount of heat in the configuration II is
A) 2. 0 s B) 3.0 s C) 4.5 s D) 6.0 s2. Ans : A
Sol : 2 1 1 2
1 2 2 1
Q kA t k At t k A
ll l
(Q, are same in both the configurations)
2013 JEEADVANCED_P1_Code::0 QPaper with & Key &Sol Solutions
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2
1
k4t 1 13kt 2 232
22t 9s 2s9
3. Two nonreactive monoatomic ideal gases have their atomic masses in the ratio 2:3.The ratio of their partial pressures, when enclosed in a vessel kept at a constanttemperature, is 4:3. The ratio of their densities is
A) 1 : 4 B) 1 : 2 C) 6 : 9 D) 8 : 9Ans : D
Sol : ART and M mN
d m
P
1 1 1
2 2 2
d m p 2 4 8d m p 3 3 9
4. A particle of mass m is projected from the ground with an intial speed 0u at an angle with the horizontal. At the highest point of its trajectory, it makes a completelyinelastic collision with another identical particle, which was thrown vertically upwardfrom the ground with the same intial speed 0u . The angle that the composite systemmakes with the horizontal immediately after the collision is
A) 4 B) 4
C) 2
D) 2
ANS : A
SOL :2 2
2 02 0
u sinv u 2gH but H2g
0u
0u
2 0v u cos v
1v u cos
H
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2 22 00 0
u sinu 2g u cos2g
If we assume that we need to find the angle made by the velocity of combined massjust after the collision
1 2
1
vtanv 4
5. A pulse of light of duration 100 ns is abosorbed completely by a small object initiallyat rest. Power of ithe pulse is 30 mW and the speed of light is 8 13 10 ms . The finalmomentum of the object isA) 17 10.3 10 kg ms B) 17 11.0 10 kg ms C) 17 13.0 10 kg ms D) 17 19.0 10 kg ms
Ans : BSol : Energy of the pulse E = pt
3 9 930 10 100 10 3 10 J
since momentum is conserved, final momentum of the object= initial momentum of pulse
9
8
E 3 10C 3 10
= 17 11.0 10 kg ms
6. In the Young’s double slit experiment using a monochromatic light of wavelength ,the path difference (in temrs of an integer n ) corresponding to any point having halfthe peak intensity is
A) 2 12
n B) 2 1
4n C) 2 1
8n D) 2 1
16n
Ans : B
Sol : 2maxmax
II I cos ,2 2
is phase difference
1cos2 2
2etc. but x2 4
2 x etc. x2 4
etc. general expression x 2n 1
4
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7. The image of an object, formed by a planoconvex lens at a distance of 8 m behindthe lens, is real and is onethird the size of the object. The wavelength of light inside
the lens is 23 times the wavelength in free space. The radius of the curved surface of
the lens is
A) 1 m B) 2 m C) 3 m D) 6 mAns : C
Sol : 1 2
1 1 1 11v u R R
v = + 8m
 V u 24m  m  u 
0
m
32
2R
1 1 3 11 08 24 2 R
R 3m
8. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to anend of another horizontal thin copper wire of length L and radius R. When thearrangement is stretched by applying forces at two ends, the ratio of the elogation inthe thin wire to that in the thick wire is
A) 0.25 B) 0.50 C) 2.00 D) 4.00Ans : CSol : Tension in both the wires is same
21 1 2
22 2 1
F / A F Ry/ AY R
l l ll
l l l l
1
2
2 1 11 4 2
ll
required ratio = 2
1
2
ll
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9. A ray of light travelling in the direction 1 ˆ ˆ32
i j is incident on a plane mirror. After
reflection, it travels along the direction 1 ˆ ˆ32
i j . The angle of incidence is
A) 30° B) 45° C) 60° D) 75°Ans : ASol : Angle between the incident ray and the reflected ray is
1 1i 3 j . i 3 ja.b 2 2cos 2ab 1 1
1 1 34
12
02 120 060
030
2 1 20
10. This diameter of a cylinder is measured using a vernier callipers with non zero error.It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of themain scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. Thediameter of the cylinder is
A) 5.112 cm B) 5.124 cm C) 5.136 cm D) 5.148 cmAns : BSol : Reading = MSR + L.C No.of divisions coincidence
= 0.055.10 2450
one division in MSL.Cn
= 5.10 + 0.024= 5.124cm
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Section2(One or More options Correct Type)
This section contains 5 multiple choice equations. Each quation has four choices (A)(B)(C) and (D) out of which ONE or MORE are correc.11. In the circuit shown in the figure, there are two parallel plate capacitors each of
capacitance C. The switch 1S is pressed first to fully charge the capacitor 1C and thenreleased. The switch 2S is then pressed to charge the capacitor 2C . After some time,
2S is released and then 3S is pressed. After some time,
02V1C
2C0V
1S 2S 3S
A) the charge on the upper plate of 1C is 02CV .
B) the charge on the upper plate of 1C is 0CV .
C) the charge on the upper plate of 2C is 0.
D) the charge on the upper plate of 2C is 0CV .
Ans : BDSol : Step : 1
potential of c1 = 2V0 andthat of c2 = 0with copper plate is positively changedStep : 2 Common polential
00
2 0
c V cV V
c cwith upper platespositively changed.Step : 3 Potential on c1 = V0
change on upper phase of c1= + cV0potential on c2 = V0 with upper plate negatively changed change on upper plate of c2= – cV0Ans : B,D
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12. A particle of mass M and positive charge Q, moving with a constant velcoity 11
ˆ4u ims ,
enters a region of uniorm static magnetic field normal to the x y plane. The region ofthe magnetic field extends from 0x to x L for all values of y . After passing throughthis region, the particle emerges on the side after 10 milliseconds with a velocity
12
ˆ ˆ2 3u i j ms . The correct statement(s) is (are)
A) The direction of the magnetic field is z derection.B) The direction of the magnetic field is z direction.
C) The magnitude of the magnetic field 50
3M
Q
units.
D) The magnitude of the magnetic field is 100
3M
Q
units.
Ans : ACSol :
Magnetic field must be along ‘–z’ direction as shown in fig.
Angle of deviation = 21
2
tan
y
x
VV
1 1tan3
6
rad
There for time sprent in the field 3 310 12 1012
T S T S
But 2 2 50
3
M M MT B
BQ TQ QAns : A,C
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13. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic
according to the equation, 1 1, 0.01 sin 62.8 cos 628y x t m m x s t . Assuming 3.14, the correct statement(s) is (are)
A) The number of nodes is 5.
B) the length of the string is 0.25 m.
C) the maximum displicement of the midpoint of the string, from its equilibriumpostion is 0.01 m
D) the fundamentals frequency is 100 Hz
Ans : BCSol : y(x,t) = 0.01 sin(62.8) cos 628 t No.of nodes is 6 since 5 loops will form.
Length of the string L = 5 5 22 2 k
5k
3.145 0.2562.8
The maximum displacement of the mid point of the string which is antinode from it’sequilibrium is 0.01Hence : B,C
14. A solid sphere of radus R and density is attched to end of a massless spring offorce constant. k . The other end of the spring is connected to another solid sphereof radius R and density 3 . The complete arrangement is placed in a liquid ofdensity 2 and is allowed to reach equillibrium. The correct statement(s) is (are)
A) the net elongation of the spring is 34
3R gk
B) the net elongation of the spring is 38
3R g
k .
C) the light sphere is paratially submerged.
D) the light sphere is completely submerged.Ans : AD
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Sol : The system is in equilibrium. Gravitational pull is greater than maximum possiblebuoyancy force hence lighter sphere is completely submerged. For lighter sphere.
Fb
mgke
ke + mg = Fb
ke+ 3 34 4R dg R 2dg3 3
34 R dg3e
k
Ans : A,D15. Two nonconduction solid spheres of radii R and 2R , haveing uniform volume
charge densities 1 and 2 respectively, touch each other. The net electric field at adistance 2R from the centre of the smaller sphere, along the line joining the centres
of the spheres, is zsero. The ratio 1
2
can be
A) 4 B) 3225
C) 3225 D) 4
Ans : BD
Sol : R 2R
For external point
331 2
2 2
4 4k R k 2R3 3
2R 5R
= 0
1 284 25
= 0
1
2
3225
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For Internal point :
331 2
32
4 4k. R k 2R .R3 3
4R 2R
1
2
4
Section3(Integer Value Correct Type)
This section contains 5 questions. The answer to each questionis a single digitinteger, ranging from 0 to 9 (both inclusive).16. A bob of mass m , suspended by a string of length 1l , is given a minimum velocity
required to complete a full circle in the vertical plane. At the highest point, it collideselastically with another bob of mass m suspended by a string of length 2l , which isinitially at rest, Both the strings are massless and inextensible. If the second bob,after collision acquires the minimum speed required to complete a full circle in the
vertical plane, the ration 1
2
ll is
Ans : 5Sol : l1Just completes vertical circular motion hence at the highest point its velocity is
1gl
In the elastic collision they exchange their velocities
1 2g 5gl l
1
2
5ll
17. A particle of mass 0.2kg is moving in one dimension under a force that delivers a
constant power 0.5W to the particle. If the initial speed 1inms of the particle is
zero, the speed 1inms after 5 s is
Ans : 5
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Sol : Power = d K.E
dt
pdt K.E‘p’ is constant
pt = 21 mv2
2pt 2 0.5 5v 5m / secm 0.2
18. The work functions of Silver and Sodium are 4.6 and 2.3eV , respectively. The ratioof the slope of the stopping potential versus frequency plot for Silver to that ofSodium is
Ans : 1Sol : shf w eV
s
whV fe e
slope = he which is independent of the nature of substance.
ratio of slopes = 1
19. A freshly prepared sample of a radioisotope of halflife 1386 s has activity 310disintegrations per second. Given that ln 2 0.693 , the fraction of the initial numberof nuclei (expressed in nearest integer percentage) that will decay in the first 80 s afterpreparation of the sample is
Ans : 4Sol : t
0N N e
No. of nuclei decay = t0 0N N N 1 e
fraction of decay = t
O
O
N 1 eN
t1 e 100% 0.041 exp 100%
1 1 0.04 100% 4%
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20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angularvelocity of 11 0 ra d s about its own axis, which is vertical. Two uniform circularrings, each of mass 6.25kg and radius 0.2m , are gently placed symmetrically onthe disc in such a manner that they are touching each other along the axis of the discand are horizontal. Assume that the friction is large enough such that the rings are atrest relative to the disc and the system rotates about the original axis. The new
angular velocity 1in rad s of the system is
Ans : 8
Sol: Apply the conservation of angular momentum
D D D ringI I ' 2I '
D D
D ring
I'I 2I
2D
22D D
ring ring
M R 1000 8125M R 2.2M R
2
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PARTII_CHEMISTRYSection1
(Only one Option correct Type)This section contains 10 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONLY ONE is correct.21. Consider the following complex ions, P, Q and R
36P FeF ,
2
2 6Q V H O
and 2
2 6R Fe H O
The correct order of the complex ions, according to their spin  only magneticmoment values (in B.M.) isA) R < Q < P B) Q < R < P C) R < P < Q D) Q < P < R
Ans : B
Sol: 36P FeF has 5 unpaired e–
2
2 6Q V H O
___ 3 unpaired e–
2
2 6R Fe H O
___ 4 unpaired e–
order Q < R < P22. The arrangement of X ions around A ion in solid AX is given in the figure (not
drawn to scale). If the radius of X is 250 pm, the radius of A is
A) 104 pm B) 125pm C) 183 pm D) 57pmAns : ASol : Fcc
2a 4r 1000pma2
= 707.21 pm
2r 2r a
2r 707.21 500.00 207.21pm
r 104pm
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23. Sulfide ores are common for the metalsA) Ag, Cu and Pb B) Ag, Cu and Sn C) Ag, Mg and Pb D) Al, Cu and Pb
Ans : ASol: Ag, Cu and Pb occur in nature as sulphide ores Ag2S, CuFeS2 or Cu2S and PbS
24. The standard enthalpies of formation of 2CO g , 2H O l and glucose (s) at 25 C
are 400 /kJ mol , 300 /kJ mol and 1300 /kJ mol respectively. The standard en
thalpy of combustion per gram of glucose at 25 C is
A) +2900 kJ B) 2900 kJ C) 16.11 kJ D) +16.11 kJAns : CSol : 2 2 1C O CO H 400 kJ .......... (1)
2 2 2 21H O H O H 300 kJ2
. ....... (2)
2 2 6 12 6 36C 6H 3O C H O H 1300 kJ ...... (3)
6 12 6 2 2 2C H O 6O 6CO 6H O ; H ?
1 6 2 6 3
per 1 mole –2900 kJper 1 gm ?Ans : –16.11 kJ
25. Upon treatment with ammoniacal 2H S , the metal ion that precipitates as a sulfide isA) Fe(III) B) Al(III) C) Mg(II) D) Zn (II)
Ans : DSol: With ammonical H2S Both Fe(III) and Zn(II) precipitate as sulphide more appro
priate is Zn (II)26. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at
25 C . For this process, the correct statement isA) The adsorption requires activation at 25 CB) The adsorption is accompanied by a decrease in enthalpyC) The adsorption increases with increase of temperature.D) The adsorption is irreversible.
Ans : BSol :Adsorption from solution is similar to adsorption of gas on solid
accompanied by decrease in enthalpy
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27. KI in acetone, undergoes 2NS reaction with each of P, Q, R and S. The rates of thereaction vary as
H3C Cl Cl Cl
Cl
O
P Q R S
A) P > Q > R > S B) S > P > R > Q C) P > R > Q > S D) R > P > S > QAns : B
Sol:
OCl
reacts with maximum rate due to participation of C = O “P”
orbitals in the stabilization of transition state i.e, SN2 T.S. chloroacetophenone is32,000 more reactive towards KI & actone at 75oC than 1  chlorobutane.
Ph
C
CH H
O Cl
Nu
T.S of “S” molecule.
Cl practically speacking allyl chloride is highly reactive than methylchloride towards SN2 r x n[Ref : Jerry March page No :  339]due to may be above reason.After allyl chloride methyl chloride is reactive and least reactive is isopropyl chloride.But with in the boundaries of given options, option “B” is best suitable one.
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28. In the reaction,P Q R S the time taken for 75% reactio of P is twice the time taken for 50% reaction of P. Theconcentration of Q varies with reaction time as shown in the figure. The overall orderof the reaction is
Q
0Q
Time
A) 2 B) 3 C) 0 D) 1Ans : D
Sol : 75%2.303 100t log
k 25
50%2.303 100 0.693t log
k 50 k
first order29. Concentrated nitric acid, upon long standing, turns yellow  brown due to the forma
tion ofA) NO B) 2NO C) 2N O D) 2 4N O
Ans : BSol: Conc HNO3 dissociate slowly
2HNO3 H2O + 2NO2 + O2Due to dissolution of NO2, Conc HNO3 upon long standing turns yellow  brown.
30. The compound that does NOT liberate 2CO , on treatment with aqueous sodiumbicarbonate solution, isA) Benzoic acid B) Benzenesulphonic acidC) Salicylic acid D) Carbolic acid (Phenol)
Ans : D
Sol:OH
+ NaHCO3 (aq) No carbondioxide liberation
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Section2(One or More Options correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE or MORE are correct.31. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is
1/100th of that of a strong acid ,1 ,HX M at 025 .C The aK of HA isA) 41 10 B) 51 10 C) 61 10 D) 31 10
Ans : A
2H 10
aH k .c
4ak 10
32. The hyperconjugative stabilities of tertbutyl cation and 2butene, respectively, aredue toA) p (empty) and * electron delocalisationsB) * and electron delocalisationsC) p (filled) and electron delocalisationsD) *p filled and * electron delocalisations
Ans : A
Sol:CH3
CH3
C+H3C & CH3 – CH = CH – CH3
stabilities can be explained as follows.In case of tertiary butyl carbocation, stabilization is due to transfer of " " electrons to empty “p” orbitals. So, the delocalization is p (empty).
CH
H
H
H
HH
HH H
C+CC
Stability of 2butene can be explained as follows.
C = C In the double bonded region two “p” orbitals of two carbons aregiving two molecular orbitals those are & * orbitals. In the hyper conjugativestabili zation of double bonded region, – bonded electrons are confined to –
molecular orbital, and C H " " bonded electrons are entering into * –mo
lecular orbital.
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33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)A) 3 25
Cr NH Cl Cl and 3 24Cr NH Cl Cl
B) 3 24Co NH Cl
and 3 22
Pt NH H O Cl
C) 22 2CoBr Cl and 2
2 2PtBr Cl
D) 3 33Pt NH NO Cl and 3 3
Pt NH Cl Br Ans : B & DSol: a
3 2 3 24 2B Co NH Cl and Pt NH H O Cl
exhibit geometrical isomers.
3 3 33 3D Pt NH NH Cl and Pt NH Cl Br exhibit ionisation isomers
34. Among P,Q,R and S, the aromatic compound (s) is/areA) P B) Q C) R D) S
3AlClP
Cl
N aHQ
R0100 115 C 4 32NH CO
O O
H C lS
O
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Ans : ABCDSol:
35. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)A) G is positive B) systemS is positive C) 0surroundingsS D) 0H
Ans : B, C, DSol : No heat absorbed or released
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Section3(Integer Value correct Type)
This section contains 5 questions. The answer to each question is a single digitinteger, ranging from 0 to 9 (both inclusive)
__________________________________________________________________36. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the
de Broglie wavelength of He gas at 073 C is “M” times that of the de Broglie wavelength of Ne at 0727 C . M is
Ans : 5
Sol :h
mv
h2m K.E
, 1 2 2
2 1 1
m Tm T
,M 20 10001 4 200
M = 537. 4EDTA is ethylenediaminetetraacetate ion. The total number of N Co O bond angles
in 1
Co EDTA complex ion is
Ans : 8
Sol:
6 are 90o
2 are 180o
Total are 8More appropriate answer = 8
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38. The total number of carboxylic acid groups in the product P is
O
O
3
32 2
1.H O ,2.O3.H O
PO
O
O
Ans : 2Sol:
39. A tetrapeptide has COOH group on alanine. This produces glycine (Gly), valine(val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this
tetrapeptide, the number of possible sequences (primary structures) with 2NH groupattached to a chiral center is
Ans : 4
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Sol : 2 2H N CH COOH Glycine
3 NH2
CH C H COOH Alanine
2 NH2
Ph CH C H COOH Phenyl Alanine
3  CH NH3 2
CH CH CH COOH Valine
For, the given tetrapeptide – COOH on alanine is fixed. This means alanine isfixed as, “C” terminus. Remaining aminoacids are glycine which is without chiralcarbon, and valine and phenyl alanine which are chiral. Then the number of possible sequences (primary structures) with  NH2 group attached to a chiral centeris four. Because we have left with only two aminoacids. Which are chiral andwhich can act as Nterminal ends. By fixing one chiral amino acid at Nterminuswe can obtain two peptide sequence by inter changing internal two aminoacids.Above phenomenon can be repeated even for second chiral aminoacid. As awhole ultimately we will be having total four peptide sequences.
40. The total number of lonepairs of electrons in melamine isAns : 6
Sol: Melamine’s structure is N N
N
NH2
H2N NH2
Total no.of lone pairs in the melamine is “6”
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PARTIII_MATHEMATICSSection1
(Only one Option correct Type)This Section contains 10 multiple choice Questions. Each question has four choices (A),(B),(C)& (D) out of which ONLY ONE is correct.
41. For 0,a b c the distance between 1,1 and the point of intersection of the lines0ax by c and 0bx ay c is less than 2 2.Then
A) 0a b c B) 0a b c C) 0a b c D) 0a b c
Ans : A (or) C
Sol :The point of intersection of two lines is c c,
a b a b
distance d < 2 2 2d 8
2c2 1 8a b
c1 2a b
a b c 0 and also a  b + c > 0 (since a > b > c > 0)42. The area enclosed by the curves sin cosy x x and cos siny x x over the interval
0,2
is
A) 4 2 1 B) 2 2 2 1 C) 2 2 1 D) 2 2 2 1
Ans : B
Sol : 1 2y 2 sin x , y 2 sin x4 4
put x4
1 23 and y 2 sin , y 2 cos
4 4
required area =
32 4
4 2
2 sin cos d 2 sin cos d
= 2 2 2 1
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43. The number of points in , , for which 2 sin cos 0x x x x , isA) 6 B) 4 C) 2 D) 0
Ans : CSol :Let 2f x x x sin x cos x f ' x x 2 cos x
y
xO
f ' x 0 x 0, f ' x 0 x 0 and f ' x 0 for x 0
44. The value of 23
1
1 1cot cot 1 2
n
n kk
is
A) 2325 B) 25
23 C) 2324 D) 24
23Ans : B
Sol :Let
23 n 231 1
n 1 k 1 n 1
n 1 ncot 1 2k tan1 n n 1
23
1 1
n 1
tan n 1 tan n
= 1tan 244
23tan25
45. A Curve passses through the point 1, .6
Let the slope of the curve at each point
,x y be sec , 0y y xx x
. Then the equation of the curve is
A) 1sin log2
y xx
B) cos log 2yec xx
C) 2s log 2yec xx
D) 2 1cos log
2y x
x
Ans : A
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Sol :dy y ysecdx x x
2
xdy ydx y dxcosx x x
y y dxcos dx x x
ysin ln x cx
but y 16
1c2
46. Let 1: ,12
f R (the set of all real numbers) be a positive, non constant and differen
tiable function such that 1 2f x f x and 1 12
f
. Then the value of 1
1/2
f x dx lies
in the interval
A) 2 1, 2e e B) 1, 2 1e e C) 1, 1
2e e
D)
10,2
e
Ans : D
Sol : 2xd e f x 0dx
2x 1e f xe
2x 1f x e
1 1
2x 1
1/2 1/2
0 f x dx e dx 1
1/2
e 10 f x dx2
47. Ler ˆˆ ˆ3 2PR i j k and ˆˆ ˆ3 4SQ i j k
determine diagonals of a parallelogram PQRSand ˆˆ ˆ2 3PT i j k
be another vector. Then the volume of the parallelepipeddetermined by the vectors PT
, PQ and PS
is
(A) 5 (B) 20 (C) 10 (D) 30
Ans : C
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Sol: From the given data PR PQ PS
but ˆ ˆ ˆPQ PS i 3j 4k
ˆ ˆ ˆ ˆ ˆ ˆPQ 2i j 3k and PS i 2 j k
volume of the parallelepiped is v PT PQ PS 10
48. Perpendiculars are drawn from ponits on the line 2 12 1 3
x y z
to the plane
3x y z . The feet of perpendiculars lie on the line
(A) 1 25 8 13x y z
(B) 1 2
2 3 5x y z
(C) 1 24 3 7x y z
(D) 1 2
2 7 5x y z
Ans : DSol :Let 2 2k, 1 k, 3k be a point on the given line and its projection on the given
plane is x 2 2k y 1 k z 3k 3 4k 3
1 1 1 3
hence the equation of required line is x y 1 z 22 7 5
49. Four persons independently solve a certain problem correclty with probabilities1 3 1 1, , , .2 4 4 8 Then the probability that the problem is solved correctly by at least one of
them is
(A) 235256 (B) 21
256 (C) 3256 (D) 253
256
Ans : A
Sol :Required propability = 1 1 3 7 2351 . . .2 4 4 8 256
50. Let complex numbers and 1
lie on circles 2 2 20 0x x y y r and
2 2 20 0 4x x y y r , respectivley. If 0 0 0z x iy satisfies the equation
2 202 2,z r then
(A) 12 (B) 1
2 (C) 17 (D) 1
3
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Ans : C
Sol :1,
lie on 0 0 z z  r and  z z  2r
respectivelty, 2
1R
(say). Here   R
0 02 z  r and z 2rR
20 0 R z  2 z
squaring and rewriting their equation
4 4 2 20 01 4R 4R 7R 2 1 7R z z 0 .......... (1)
But by hypohesis we have2 2
02  z  r 2
0 0 0 02z z z z 2 2
0 0 0 0z z R z z 2 ...... (2)substituting (2) in (1)
20 01 7R 1 z z 0
1R7
Section2(One or More Options Correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C)and (D) out of which ONE or MORE are correct.51. A line l passing through the origin is perpendicular to the lines
1ˆˆ ˆ: 3 1 2 4 2 ,l t i t j t k t
2ˆˆ ˆ: 3 2 3 2 2 ,l s i s j s k s
Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point ofintersection of l and l1 is (are)
A) 7 7 5, ,3 3 3
B) 1, 1,0 C) (1,1,1) D) 7 7 8, ,9 9 9
Ans : BD
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Sol :Let line passing through the origin and perpendicular to given lines is x y
mz
n
l
2m 2n 0 l
2 2m n 0 l2m 2n 0 m n
2 2m n 0 2 3 2
l ll
equation of the line l is x y z2 3 2
and we have line 1x 3 y 1 z 4is
1 2 2
l
Point of intersection of 1l and l is (2, –3, 2)
By given condition, 17 is distance between (2s + 3, 2s + 3, s + 2) and (2, –3, 2)29s 28s 20 0
10s 2, s9
52. Let sin , 0f x x x x . Then for all natural numbers n, ( )f x vanishes at
A) a unique point in the interval 1,2
n n
B) a unique point in the interval 1 , 1n n2
C) a unique point in the interval (n,n+1)
D) two points in the interval (n,n+1)Ans : BC
Sol : 1 sin x 1 x x sin x x and 1f n 0, n N2
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53. Let 1
4 2
11
k kn
nk
S
k2. Then Sn can take value(s)
A) 1056 B) 1088 C) 1120 D) 1332Ans : ADSol : 2 2 2 2 2 2 2 2
nS 1 2 3 4 5 6 7 8 . ..........upto 4n terms
2 2 2 2 2 2 2 2 2 23 1 7 5 11 9 ........... 4 2 8 6 .....
2 3 1 7 5 ..... 4 2 1 4 3 . ....
2 2 28n 8n 4n 16n 4n
ns 1056 for n 8 and 1332 for n 9
54. For 3 3 matrices M and N, which of the following statement(s) is (are) NOTcorrect?
A) NTMN is symmetric or skew symmetric, according as M is symmetric or skewsymmetric
B) MNNM is skew symmetric for all symmetric matrices M and N
C) M N is symmetric for all symmetric matrices M and N
D) (adj M) (adj N) = adj (M N) for all invertible matrices M and NAns : CD
(A)
Sol :If TT T T T TM M then N MN N M N N MN
If TT T T T TM M then N MN N M N N MN
(B)
If T T TT TM M and N N then MN NM MN NM
= T T T TN M M N N M MN MN NM
(C)
If T TM M and N N then T T TMN N M NM
(D)Adj (MN) = adj (N) adj (M)
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55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:15is converted into an open rectangular box by folding after removing squares of equalarea from all four corners. If the total area of removed squares is 100, the resultingbox has maximum volume. Then the lengths of the sides of the rectangular sheet areA) 24 B) 32 C) 45 D) 60
Ans: AC(Approximately)Sol: Question not clear
Section3(Integer Value correct Type)
This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9 (both inclusive)
56. Consider the set of eight vectors ˆˆ ˆ : , , 1,1V ai bj ck a b c .Three non coplanarvectors can be chosen from V in 2P ways. Then p is.
Ans : 5Sol :If a, b, c 1, 1, 1
the no. of ways in which 3 vectors are coplanar is 24
1, 1, 1 , 1, 1, 1 . .....6ways, 1, 1, 1 , 1, 1, 1 .....6 ways
1, 1, 1 , 1, 1, 1 . .....6ways, 1, 1, 1 , 1, 1, 1 .....6 ways
but total ways = 56 noncoplanar vectors = 32
57. Of the three independent events 1E , 2E and 3E the probability that only E1 occurs is , only E2 occurs is and only E3 occurs is . Let the probability p that none ofevents E1,E2 or E3 occurs satisfy the equations 2 3 2p and p .All the given probabilities are assumed to lie in the interval (0,1).
Then 1
3
Probability of occurrence of EProbability of occurrence of E
Ans : 6
58. The coefficients of three consecutive terms of 51 nx are in the ratio 5:10:14.Then n =
Ans : 6
Sol: 5 24 15 14
n 5 11 n 6100 70
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59. A pack contains n cards numbered from 1 to n. Two consecutive numbered cardsare removed from the pack and the sum of the numbers on the remaining cards is1224. If the smaller of the numbers on the removed cards is k, then k20 =
Ans : 5Sol :Let the removed numbers are k, k + 1
2n n 11224 2k 1 n n 4k 2450
2
by observation n = 50, k = 25
60. A vertical line passing through the point (h,0) intersects the ellipse 2 2
14 3x y
at the
points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If
h = area of the triangle PQR, 1 2 1/2 11/2 1max min
hhh and h
, then 1 2
8 85
Ans : 9Sol :Here the tangents at P and Q to the ellipse are intersect at Q on its major axis
area 23 4 hh 1
2 h 4
here is decreasing from 1 to 12
1 245 95,8 2
1 28 8 95