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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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JEE Advanced Paper -2 Question Paper with Solutions and Key


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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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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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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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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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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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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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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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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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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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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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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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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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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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:

(A) SmartPrep.in



(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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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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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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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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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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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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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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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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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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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

the following is/are true?

A) 3 tan tan 02 2

B) tan 3 tan 02 2

C) 3 tan tan 02 2

D) tan 3 tan 02 2

Key : B,D

Sol : 2 cos cos cos cos 1

2cos cos cos 1 2 cos

cos 2 cos 1 2cos

1 2coscos2 cos

cos 1 1 2cos 2 coscos 1 1 2cos 2 cos

2 22

2 2

2cos / 2 3 3cos 3.2cos / 2cos / 22sin / 2 cos 1 2sin / 2

2 2tan / 2 3 tan / 2

tan 3 tan2 2

(i) tan 3 tan2 2 (ii) tan 3

2 2tn

tan 3 tan 02 2 & tan 3 0

2 2tn

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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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54. 12a

A) 11 102a a B) 11 10a a C) 11 102a a D) 11 10a aKey : D

Sol :

12 12 10 10

12 1011 11

11

p pa a p qa p

10 2 10 2

11 11

1 1

10 10

2 211 11 1 ; 1x

=1

12 10 11 12 10 11a a a a a a

* * * * *

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