21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 3 PART II : PHYSICS SECTION 1 (Maximum Marks:21) This section contians SEVEN questions Each question has FOUR options [A],[B],[C] and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct option is darkened Zero Marks : 0 If one of the bubbles is darkened Negative Marks : -1 In all other cases 01. Three vectors , P Q and R are shown in the figure let s be any point on the vector R .The distance between the points P and S is b R The general relation among vectors , PQ and S is A) 2 1 S bP bQ B) 1 S bP bQ C) 2 1 S b P bQ D) 1 S b P bQ Key : B Sol : S P bR from diagram Given R Q P S P bQ P 1 S bP bQ 02. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3x10s times heavier than the Earth and is at a distance 2.5xL04 times larger than the radius of the Earth. The escape velocity from Earth’s gravitational field is ve : 17.2 km s-1. The minimum initial velociU (us) required for the rocket to be able to leave the Sun-Earth system is closest to Ignore (the rotation and revolution of the Earth and the presence of any other planet) A) 1 72 s v kms B) 1 22 s v kms C) 1 62 s v kms D) 1 42 s v kms Key : A Sol : By conservation of energy SmartPrep.in Downloaded from http://smartprep.in
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 3
PART II : PHYSICSSECTION 1 (Maximum Marks:21)
This section contians SEVEN questions Each question has FOUR options [A],[B],[C] and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkenedZero Marks : 0 If one of the bubbles is darkenedNegative Marks : -1 In all other cases
01. Three vectors ,P Q and R
are shown in the figure let s be any point on the vector R .The
distance between the points P and S is b R
The general relation among vectors ,P Q
and S is
A) 21S b P b Q
B) 1S b P b Q
C) 21S b P b Q
D) 1S b P b Q
Key : B
Sol : S P bR from diagram
Given R Q P
S P b Q P
1S b P bQ
02. A rocket is launched normal to the surface of the Earth, away from the Sun, along the linejoining the Sun and the Earth. The Sun is 3x10s times heavier than the Earth and is at adistance 2.5xL04 times larger than the radius of the Earth. The escape velocity from Earth’s
gravitational field is ve : 17.2 km s-1. The minimum initial velociU (us) required for therocket to be able to leave the Sun-Earth system is closest toIgnore
(the rotation and revolution of the Earth and the presence of any other planet)A) 172sv kms B) 122sv kms C) 162sv kms D) 142sv kmsKey : ASol : By conservation of energy
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MEDHA-TEJA
Typewritten text
JEE Advanced Paper -2 Question Paper with Solutions and Key
21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 4
212
s Ee
E
GM m GM m mVr R
Ms = mass of sunr = distance between sun and earth= 2.5 X 104 RERE = radius of earthME = mass of earthm= mass of earthSubstitutiog the given values and simpligyingVe = 72 km/s
03. A person measures the depth of a well by measuring the time interval between dropping astone and receiving the sound of impact with the bottom of the well. The error in hismeasurement of time is dI L is / = 0.01 seconds and he measures the depth of the well to beL= 20 rneters. Take the acceleration due to gravity g = 0 ms-2 and the velocity of sound 300
2ms Then the fractional error in the measurement, 6L/L, is closest toA) 1% B) 2% C) 5% D) 3 %Key: A
Sol : 2L Ltg V
Differentiation
2 1 1.2
t L Lg VL
Dividig b L
2 1 1. .2
t L LL g L V LL
12 1 1.
2L t
L L g VL
Substitution the values
1%LL
04 Conside an expanding sphere of instantaneous radius R whose total mass remains constant’The expansion is such that the instantaneous density remains uniform throughout the
volume rate of fractional change in density 1dppdt
is constant. The velocity v of any point on
the surface of the exanding sphete is propotional to
A) R B) 3R C) 1R D) 2 /3R
Key: A
Sol: 3483
M R
M = constant Differentation
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 5
0- 343
p R
+ 24 33
P R R simplifying
3 pR rp
3 R pRt p t
tanp cons tp t
R Rt
05 A photoelectric material having work function 0 is illuminated with light of wavelength
0
hc
The fastest photoelectron has a do Broglie wavelength .d A change in
wavelength of the incident light by results in a changge d din Then the ratio
/d d is proportional to
A) 2 2/d B) /a C) 3 2/a D) 3 /a Key : C
Sol : 2
20 2
hc hc hm d
Differentiation
2
hc
2
3
22 d
h dm
3
2
d d
06 Consider regular polygons with number of sides n = 3,4,5 .........as shown in the figure The
center of mass of all the polygons is at height h from the ground. They roll on a horizontalsurface bout the lending vertex without slipping and sliding as depicted. The maximumincrease in height of the locus of the center of mass for cach polygon is Then depends onn and h as
A) 1 1
cosh
n
B) 2sinh
n
C) 2tan
2h
n
D)
2sinhn
Key : A
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 6
Sol : (Triangle)For n = 3h = 2h
2h h h for n = 4
2 2 1h h h h h
Theefore, conect oftian is [A]07 A symetric stat shaped conducting wire loop is carrying a Steady state current / as shown in
the figure. The distance between the diametrically opposite vertices of the star is 4a Themagnitude of the magnetic field at the center of the loop is
A) 0 3 3 14
la
B) 0 3 2 34
la
C) 0 6 3 14
la
D) 0 6 3 14
la
Key : DSol : let us take a segment of each triangle of stage whosePerpendicular distance from the centre is‘a’ from data given in figure and its ends subtends angle 600 and 300 with the centre
B of each segment is 0 sin 60 sin 304s
iBa
0 1 3 14 2s
iBa
Total segments are 12
0 112 3 14 2net
iBa
But = 0 6 3 14
iButa
SECTION 2 (Maximum Marks : 28) This section contians SEVEN questions Each question has FOUR options [A],[B],[C] and [D]. ONE OR MORE THAN ONE of these
four options is (are ) correct For each question,darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories
Full Marks: +4 If only the bubbles (s) corresponding to all the correct option(s) is(are) darkenedPartial Marks : +1 For darkening a bubble corresponding to each correct option,
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 7
provided NO incorrect option is darkenedZero Marks : 0 If none of the bubbles is darkenedNegative Marks: -2 In all other cases
For example, if [A],[C] and [D] are all the correct options for a question, darkening all these threewill get +4 marks; darkening only [A] and [D] will get +2 marks; and darkening [A] and [B] will get -2 marks, as a wrong option is also darkened
8) A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor(as shown in the figure) At some instant of time, the angle made by the har with the vertical is Which of the following statements about its motion is are correct ?
A) When the bar makes an angle with the vertical, the displaccment of its midpoint from the initialposition is proportian) to ( 1-cos )B) The midpoint of the bar will fall vertically downwardC) The trajectory of the point A is a parabolaD) Instantancous torque about the point in contact with the floor is proportin contact with the floor isproportional to sin Key : A,B,D
Sol : a Displacement of centre is cos2ls
b) path of centre of mass is a striaght linec) It is the edge of rodd) torgue is proportinal to perpendicular distance which is proportianal to sin Sm
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 8
9) A source of constant voltage V is connected to a reststance R and two ident inductors L1andL2 through a switch S as shown There is no mutual inductance between the two inductors Theswitch S is initially open at t=0 the swich is ctosed and current begins to flow wich of thefollowing options is/are correct ?
A) after n long time, the current through L2 will be 1
1 2
v LR L L
B) The ratio of the currents through L1 and L2 is fixed at all times (t > 0)
C) After a long time the current through L1 will be 2
1 2
v LR L L
D) at t = 0 the current through the resistance R is vR
Key : A,B,CSol : 1 2v v
1 1 2 2p L l L
1 2
2 1
p ll L
1 2
2 1
l L Ll L
1 1
21 2 1 2
L V Ll iL L R L L
Current through 1
2 21 2
v LL is lR L L
1 2
2 1
l LL L
Options A,B & C current
10. A point charge Q is placed just outside an imginary hemispherical surface of radius R asshown in the figure. Which of the the following statements is/are correct
A) Total flux through the curved and the flat surfaces is 0
Q
B) The circumference of the surface is an equipotentialC) The component of the electric field normal to the flat surface is constant over the surface
D) The electric flux passing through the curved surface of the hemisphere is 0
112 2Q
Key : B,DSol : All points on circumference are at equal distance from point change Hence polential is same at allpoints on circumference of flat surface,Option ‘B’ is correct
0
1 cos2totalq
0
12 2q R
R
q
2RQR
R
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 9
0
112 2totalqQ
Option D also correct B & D are correct11. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in
the figure. A constant force is continously applied on the surface of the wheel so that it justclimbs the step without slipping. Consider the torgue r about an axis normal to the plane of thepaper of the paper passing through the point Q. Which of the following options is/are correct?
A) If the force is applied normal to the circumfernce at point X then r is constantB) If the force is applied at point P tangentially then r decreases continously as the wheel climbsC) If the force is applied tangentially at point S then r 0 but the wheel never climbs the stepD) If the force is applied normal to the circumference at point P then r is zeroKey : B,DSol : Conceptual
12. A uniform magnetic field B exists in the region between 0x and 32Rx (region 2 in the
figure) pointing normally into the plane of the paper. A particle with charge +Q and Which ofthe following option(s) is/are correct?
A) For 8 ,
13PB
QR the particle will enter region 3 through the point 2P on x axis
B) When the particles re-enters region 1 through the longest possible path in region 2, the magnitudeof the chagne in its linear momentum between point 1P and the fartherst point from y-axis is / 2p
C) For a fixed B, particles of same change Q and same velocity v, the distance between the point 1P andthe point of re-entry into region 1 is inversely proportional to the mass of the particle
D) For 2 ,3
PBQR
the particle will re-enter region 1
Key : A,D
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 10
Sol : 23
mv mvr PBq Qqr
32
r
13. Two coherent monochromatic point sources 1 2S and S of wavelength 600 nm are placedsymmetrically on either side of the center of the circle as shown. THe sources are separated bya alternate bright and dark spots on the circumference of the circle. THe angular separationbetween two cosecutive bright sports is . Which of the following options is/are correct?
A) A dark spot will be formed at the point zPB) At zP the order of the fringe will be maximum
C) The angular separetion between two consecutive bright spots decreases as we move from 1 2P to P alongthe first quadrantD) The total number of fringes produced between 1 2P and P in the first quadrant is close to 3000Key : B,DSol : cosd n
9
3
600 10cos1.8 10
n nd
nearly no. of fringes = 3000
14. The instantaneous voltages terminals marked X,Y and Z are are given by
0 sin ,xV V t
02sin3YV V t and
04sin3zV V t
An ideal voltmeter is configured to read rms value of teh potential difference between terminalsIt is connected between points X and Y and then between Y and Z. The reading (S) of the voltmeterwill be
A) 012
rmsXYV V B) independent of the choice of the two terminals
C) 012
rmsYZV V D) 0
rmsYZV V
Key : CSol : x y z
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 11
2 2 20 0 0 0
22 cos3xyV V V V V
0
2rms
xyVV
0
2rms
yzVV
correct optiond is ‘c’SECTION 3 (Maximum Mars: 12)
This section contains TWO questions Based on each paragraph, there are TWO questions. Each questions has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four op-
tions is correct. For each question, darken the bubble corresponding to the correct integer in the ORS For each question, marks will be awarded in one of the following categories:
Full marks : +3 If only the bubble corresponding to the correct answer is darkenedZero Marks : 0 In all other cases
15. In process 1, the energy stored in the capacitor CE and heat dissipated across resistance DE .are related by:
A) 2C DE E in B) 2C DE E C) C DE E D) 12C DE E
Key : CSol : C DE E Energy is capacitor = Energy disiplated in resistor
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 12
16. In process 2, total energy dissipated across the resistance DE is:
A) 20
12DE CV B) 2
03DE CV C) 2
01 12 2DE CV
D) 2
0132DE CV
Key : A
Sol : 2 2 2 20 2 1 0 3 2
1 12 2
U C V V C V V 20 0
12
C V
17. The total kinetic energy of the ring is
A) 20
12
M R r B) 220
32
M R r C) 220M R r D) 2 2
0M R
Key : C
Sol : M.I of ring ring = 22M R r
18. The minimum value of 0 below which the ring will drop down is
A) 2gR r B)
32
gR r C)
gR r D) 2
gR r
Key : C
Sol : 20MrW mg
gR r
1r R r
PART - II CHEMISTRYSECTION 1 (Maximum Marks:21)
This section contians SEVEN questions
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 13
Each question has FOUR options [A],[B],[C] and [D]. ONLY ONE of these four options iscorrect
For each question, darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkenedZero Marks : 0 If one of the bubbles is darkenedNegative Marks : -1 In all other cases
19. For the following cell,
4 4Zn s ZnSO aq CuSO aq Cu s
When the concentration of 2Zn is 10 times the concentration of 2Cu , the expression for G (in1Jmol ) is
[F is Faraday constant, R is gas constant, T is temperature, 0E (cell)= 1.1 V)A) 2.303 2.2RT F B) 2.2F C) 1.1F D) 2.303 1.1RT FKey : A
Sol : 0 logeG G RT Q 2
2
ZnQ
Cu
20
10 22.303 logcell
ZnG nFE RT
Cu
10102 1.1 2.303 log1
G F RT
2.2 2.303G F RT
2.303 2.2G RT F 20. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water
changes the freezing point of the solution. Use the freezing point depression constant for wateras 2 K kg mol-1. The figure shown below represent plots of vapour pressure (V.P) versustemperature (T). (Molecular weight of ethanol is 46 g mol-1]Among the following, the option representing change in the freezing point is
A) B)
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 14
C) D)
Key: C
Sol: 1000.f f
wT km W
34.5 10002 346 500
21. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 Kare
10f G C graphite kJ mol
12.9f G C diamond kJ mol The standard state means that the pressure should be 1 bar, and substance should be pure at agiven temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reducesits volume by 6 3 12 10 m mol . If C(graphite) is converted to C(diamond) isothermally at T=298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information: 2 21 1 ,J kg m s 1 21 1 ,Pa kg m s 51 10bar Pa ]A) 29001 bar B) 58001 bar C) 14501 bar D) 1450 barKey: CSol: 0 10G graphite kJ mol Graphite Diamond
0 12.9G diamond kJ mol 102.303 logG RT k 3
102.9 10 1.303 0.987 298 logk
10log 4.22k 4.22 410 1.45 10 14501K bar
22. The order of basicity among the following compounds is
A) II > I > IV > III B) IV > I > II > III C) I > IV > III > II D) IV > II > III > IKey: BSol: Conceptual
23. Which of the following combinations will produce 2H gas?
A) Au metal and NaCN (aq) in the presence of air B) Fe metal and conc. 3HNO
C) Zn metal and NaOH(aq) D) Cu metal and conc. 3HNO
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 15
Key: C
Sol: 2 2 22 sZn NaOH Na ZnO H
24. The order of the oxidation state of the phosphorus atom in 3 2 3 4 3 3, ,H PO H PO H PO and 4 2 6H P O is
A) 3 4 3 2 3 3 4 2 6H PO H PO H PO H P O B) 3 3 3 2 3 4 4 2 6H PO H PO H PO H P O
C) 3 4 4 2 6 3 3 3 2H PO H P O H PO H PO D) 3 2 3 3 4 2 6 3 4H PO H PO H P O H PO Key: CSol: Conceptual
25. The major product of the following reaction is
A) B)
C) D)
Key: CSol: Sm
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 16
26. Among the following, the correct statement (s) is (are)
A) 3 3Al CH has the three-centre two-eelectron bonds in its dimetric structure
B) 3AlCl has the three - centre two - electron bonds in its dimetritc structure
C) 3BH has the three - centre two - electron bonds in its dimetritc structure
D) The Lewis acidity of 3BCl is greater than that of 3AlClKey:A, C, DSol: Conceptual
27. For a reaction taking place in a container in equilibrium with its surroundings, the effect oftemperature on its equilibrium constant. K in terms of change in entropy is described byA) With increase in temperature, the value of K for exothermic reaction decreases because favourablechange in entropy of the surroundings decreasesB) With increase in temperature, the value of K for endothermic reaction increases because the entropychange of the system is negativeC) With increase in temperature, the value of K for endothermic reaction increases because unfavourablechange in entropy of the surroundings decreasesD) With increase in temperature, the value of K for exothermic reaction decreases because the entropychange of the system is postiveKey: A, CSol: Conceptual
28. The option(s) with only amphoteric oxides is (are)A) 2 3 2, , ,Cr O BeO SnO SnO B) 2 3 2, , ,ZnO Al O PbO PbO
C) 2 3 2, , ,NO B O PbO SnO D) 2 3, , ,Cr O CrO SnO PbOKey: A, BSol: Conceptual
29. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. Thecorrect option(s) among the following is (are)A) The activation energy of the reaction is unaffected by the value of the steric factorB) The value of frequency factor predicted by Arrhenius equation is higher than that determinedexperimentallyC) Since P= 4.5, the reaction will not proceed unless an effective catalyst is usedD) Experimentally determined value of frequency factor is higher than that prediced by Arrhenius equationKey: A, BSol: Conceptual
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 17
30. For the followingt compounds, the correct statement(s) with respect to nucleophilic substitutionreactions is (are)
A) Compound IV undergoes inversion of configurationB) The oreder of reactivity for I, III and IV is IV>I>IIIC) I and III follow 1NS mechanism
D) I and II follow 2NS mechanismKey:A, B, C, DSol: Conceptual
31. Compounds P and R upon ozonolysis produce Q and S respectively. The molecular formula ofQ and S is 8 8C H O . Q undergoes Cannizzaro reaction but not haloform reaction, whereas Sundergoes haloform reaction but not Cannizzaro reaction
i) 3 2 2
28 8
) /) /
i O CH Clii Zn H O
C H OP Q
ii) 3 2 2
2 8 8
) /) /
i O CH Clii Zn H O C H O
R S
A) B)
C) D)
Key: A, BSol:
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 18
(B)
32. The correct statement(s) about surface properties is(are)A) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption ofehtane will be more than that of nitrogen on same amount of activated charcoal at a given temperatureB) Adsoprtion is accompained by decrease in enthalpy and decrease in entropy of the systemC) Brownian motion of colloidal particles does not depend on the size of the particles but dependson viscosity of the solutionD) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion mediumKey: A, B, DSol: Conceptual
PARAGRAPH 1Upon heating 3KClO in the presence of catalytic amount of 2MnO , a gas W is formed. Excess amount of
W reacts with white phosphorus to give X. The reaction of X with pure 3HNO gives Y and z33. W and X are respectively
A) 2O and 4 6P O B) 3O and 4 10P O C) 3O and 4 6P O D) 2O and 4 10P OKey: A
Sol:
23 22 2 3MnO
gHeatKClO KCl O
W
34
4 2 4 10 3 2 55 4HNOP O P O HPO N OW X Z Y
34. Y and Z are respectivelyA) 2 4 3N O and HPO B) 2 3 3 4N O and H PO C) 2 5 3N O and HPO D) 2 4 3 3N O and H POKey: C
PRAGRAPH 2
The reaction of compound P with 3CH MgBr excess in 2 5 2C H O followed by addition of 2H O
gives Q. The compound Q on treatment with 2 4H SO at 0°C gives R. The reaction of R with
3CH COCl in the presence of anhydrous 3AlCl in 2 2CH Cl followed by treatment with
2H O produces compound S. [Et in compound P is ethyl group]
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 19
35. The reactions, Q to R and R to S, areA) Dehydration and Friedel - Crafts acylationB) Friedel - Crafts alkylation, dehydration and Friedel - Crafts acylationC) Aromatic sulfonation and Friedel - Crafts acylationD) Friedel - Crafts alkylation and Friedel - Crafts acylation.Key: DSol: Conceptual
36. The product S is
A) B)
C) D)
Key: BSol:
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 20
PART-III : MATHEMATICSSECTION 1(Maximum Marks:21)
This section contians SEVEN questions Each question has FOUR options [A],[B],[C] and [D]. ONLY ONE of these four options is correct For each question,darken the bubble corresponding to the correct option in the ORS For each question, marks will be awarded in one of the following categories
Full Marks : +3 If only the bubble corresponding to the correct option is darkenedZero Marks : 0 If none of the bubbles is darkenedNegative Marks: -1 In all other cases
37. The equation of the plan epassing through the point (1,1,1) and perpendicular to the planes2 2 5x y z and 3 6 2 7x y z ,is
A) 14 2 15 3x y z B) 14 2 15 31x y z
C) 14 2 15 1x y z D) 14 2 15 27x y z Key : BSol : 14(2) 1(2) 2(15) 0
3(14) 2( 6) 15( 2) 0 (1,1,1) lies on 14x+2y+15z=31
38. Let {1,2,3...,9}S .For k=1,2,...,5.let kN be the number of subsets of S, each containing five
elements out of which exactly k are odd. Then 1 2 3 4 5N N N N N A) 252 B) 125 C) 126 D) 210Key : C
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 21
Sol : 1, 2,3,.....,9S
1, 2,.....,5k
kN no. of subsets of S
1 4 2 3 3 2 4 1 55 .4 5 .4 5 .4 5 .4 5c c c c c c c c c
5 10 4 10 6 5 4 1 5 40 60 20 1 126
39. If :f is a twice differentiable function such that ''( ) 0f x for all x , and
1 1 , (1) 1,2 2
f f
then
A) '(1) 1f B) ' (1) 0f C) '1 (1) 12
f D) ' 10 (1)2
f
Key : A
Sol : 2f x x ax b
11 112 0f x satisfiesf x
1 1 1 1 1 2 4 22 2 4 2 2
af b a b
2 4 1a b
1 1 1 1 0f a b a b b a
11 2 12
a a
12
b
Now 2 11 1 122 2 4
f x x x f x x
1 131 1 12
f ie f
40. Let O be the origin and let PQR be an arbitary triangle. The point S is such that
. . . . . .OP OQ OR OS OR OP OQ OS OQ OR OP OS
. Then the traianlge PQR has S as its
A) Circumcentre B) incentre C) centroid D) orthocenterKey : D
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 22
Sol :
RQ
P
. . . .OP OQ OR OS OR OP OQ OS
. .OP OQ OR OS OQ OR
. .OP RQ OS RQ
. 0RQ OP OS
. 0RQ SP S is the orthocentie
41. How many 3 3 matrices M with entries from {0,1,2} are there, for which the sum of thediagonal entries of TM M is 5?A) 198 B) 135 C) 126 D) 162Key : A
Sol : Let
a b cm d e f
g h i
2 2 2 2 2 2 2 2 5TM M a b c e f g h i where entries are {0,1,2}case (i) five ethlies and other free zero
51Ca
case (ii) one side ‘2’ one etly is ‘1’ and other zero
22!Ca
total =126+72=198
42. If y =y(x) satisfies the differential equation 1
9 4 9 , 0B x x dy x dx x
and
y(0)= 7, then y(256)=A) 80 B) 9 C) 16 D) 3Key : DSol : 29 x t
24 9dx t t dt
G.E
2 218 9 94
t t dy t dtt
124
dy dtt
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 23
124
dy dtt
2 2 4y t c
42cy t
4 9 (1)2cy x
(0) 7y
7 7 02c c
(256) 4 9 256 0 3y 43. Three randomly chosen nonnegative integers x,y and z are found to satisfy the equation
x+y+z=10. Then the probability that z is even is
A) 5
11 B) 12 C)
611 D)
3655
Key : CSol :
3 1 2( ) 10 3 1 12C Cn S
z=then =2k10 2x y k where k=0,1,2,3,4,5
2 1 1
10 2 2 1 1 2 11 2C Ck k k
5
0( ) 11 2 11 9 7 5 3 1
kn E k
36
2
( ) 36 36 2( )( ) 12 12 11C
n EP En S
611
SECTION 2 (Maximum Marks : 28)* This section contains SEVEN questions* Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these
four option(s) is(are) correct.* For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.* For each question, marks will be awarded in one of the following categories :
Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened
Zero Marks : 0 If none of the bubbles is darkenedNegative Marks : -2 In all other cases
* For example, if (A), (C) and (D) are all the correct options for a question, darkening all these
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 24
three will result in +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A)and (B) will get -2 marks, as a wrong option is also darkened.
44. If the line x= divides the area of region 2 3[( , ) . 1]R x y x y x into two equal parts,then
A) 102
B) 4 22 4 1 0 C) 4 24 1 0 D) 1 12
Key : C,D
Sol :
X
y n3y n
P
0, 0
1Y
1X
Y
0
1,1
x
By data 1
3 3 4 22 4 1 0D
x x dx x x dx
4 2 12 02
22 112
2 112
2 1 2 1,2 2
But 2 1
2
(Not Possible),
2 12
(Satisfies 1 12
)
45. If :f is a differentiable function such that '( ) 2 ( )f x f x for all ,x and f(0)=1, then
A) 2( ) (0, )xf x e in B) ' 2( ) (0, )xf x e in
C) ( )f x is decreasing in (0, ) D) ( )f x is increasing in (0, )Key : A,DSol : Given ': , ( ) 2 ( ),f R R f x f x x R and
(0) 1f
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 25
Now '( ) 2( )
f xf x
'( ) 2( )
f x dx dxf x
log ( ) 2e f x x c
2( ) .xf x e Abut f(0)=1
0(0) .f e A
1 1A A but 0cA e
0 1A 2( . ) ( ) . xi e f x A e
2. ( )xA e f x
but 2 2. x xA e e20 ( ) xf x e
2 ( )xe f x2( ) xf x e
Now 2( ) . xf x A e2( ) 2 . 0xf x A e
( )f x is increasing in (0, )
46. Let 1 (1 1 ) 1( ) cos 1
1 1x x
f x for xx x
.Then
A) 1lim ( )x f x does not exist B) 1lim ( )x f x does not exist
C) 1lim ( ) 0x f x D) 1lim ( ) 0x f x Key : A,D
Sol : lim lim
1 1
1 (1 1 ) 1( ) cos1 1x x
x xf x
x x
lim1
1 (1 (1 )) 1cos(1 ) 1x
x xx x
=does not exist
lim lim1 1
1 1 1 1( ) cos1 1x x
x xf x
x x
lim1
1 (1 (1 )) 1cos(1 ) 1x
x xx x
lim1
11 cos (0) ( 1 1)1x x finitevaluebetween and
x
=0
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 26
47. If sin(2 ) 1
sin( ) sin ( )
x
xg x t dt , then
A) ' 2
2g
B) ' 2
2g
C) ' 2
2g
D) ' 2
2g
Key : A,C
Sol : sin2
1
sin( ) | sin ( )
x
xg x t dt
'( ) (2 )(2cos 2 ) .cosg x x x x x ' 4 ( 1) 2
2 2g
' 4 ( 1) 22 2
g
48. If 198
1
1( 1)
k
k k
kI dxx x
, then
A) 4950
I B) 4950
I C) log 99eI D) log 99eI
Key : A,D
Sol : 1 1 11 1 1
1 1 1kx k
x k x
11 1 1 11 ( 1)
k
k
k k dx dxx x x x x x
198 98
1 1
1 1log( 1)
k
k kk
k kdxx x k
log(99)I
Also 1 1 11 1 100 100
k kx x k
1 1 1( 1) 100
k
k
k dxx x
198 98
1 1
1 1( 1) 100
k
k kk
k dxx x
1 4998100 50
I I
49. If
cos(2 ) cos(2 ) sin(2 )( ) cos cos sin ,
sin sin cos
x x xf x x x x
x x x then
A) '( )f x =0 at exactly three points in ,
B) '( ) 0f x ( , )
C) ( )f x attains its maximum at x=0D) f(x) attains its minimum at x=0
Key : A,C
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 27
Sol : cos 2 cos 2 sin 2
cos cos sinsin sin cos
x x xf x x n x
x x x
1 1 2C C C
0 cos 2 sin 2
2cos cos sin0 sin cos
x xf x x x x
x x
20 cos 2 2 cos 0 sin 2 2sin cos 0x x x x x
2 24cos 1 sin 2x x
1 6sin 2f x x
1 0 sin 2 0 0f x x in x in , at 20, ,2 2
x A & f x altains maximum at
0x 50. Let and be nonzero real numbers such tath 2 cos cos cos cos 1. The which of
21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 28
SECTION 3 (Maximum Marks: 12)* This section contains TWO paragraphs.* Based on each paragraph, there are TWO questions.* Each question has Four options (A),(B),(C)and (D).ONLY ONE of these four options is correct.* For each question,darken the bubble corresponding to the correct option in the ORS.* For each question,marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 In all other cases.
51. OX OY
A) sin Q R B) sin 2R C) sin( )p R D) sin p QKey : D
Sol :
OX
OYOZ
RQ
P
Given 1OX OY OZ
OX OY OX OY SIN XOY
1.1.sin R
sin 180 P Q
sin .P Q D
52. If the triangle PQR varies, then the minimum value of 'cos( ) cos( ) cos( )P Q Q R R P is
A) 53
B) 32
C) 53 D)
32
Key : B
Sol : Cos P Q Cos Q R Cos R P
CosR CosP CosQ to have the minimum 060P Q R 32
53. If 4 28,a then p+2q=A) 12 B) 14 C) 21 D) 7Key : A
Sol :
4 4
4 44
1 5 1 5 56 24 528 28 ( ) ( )2 4 16 16
a p q p q p q p q
5628 ,016
p q p q 4p q and thus 2 12p q
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21 May 2017 JEE ADVANCED 2017 P2 QUESTION PAPER (CODE-6) 29
54. 12a
A) 11 102a a B) 11 10a a C) 11 102a a D) 11 10a aKey : D