JEE ADVANCED (PAPER 2) SOLUTIONS PART I : PHYSICS Advanced Pap… · JEE ADVANCED (PAPER – 2) SOLUTIONS ... M d 4 dR. .3R . dt 3 dt ... S 1 b P bQ (C) S 1 b P b Q 2 (D) S b 1 P

JEE ADVANCED (PAPER – 2) SOLUTIONS – 2017 (Code-3)

IIT KALRASHUKLA– CIVIL LINES, KAKADEO, SOUTH KANPUR – 7617 800 800 [1]

PART I : PHYSICS

SECTION 1 (Maximum Marks : 21)

• This section contains 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 none of the bubbles is darkened.

Negative Marks : –1 In all other cases.

1. 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 3 105 times heavier than the Earth and is at a distance

2.5 104 times larger than the radius of the Earth. The escape velocity from Earth’s gravitational

field is ve = 11.2 km s–1. The minimum initial velocity (vs) 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) vs = 22 km s–1 (B) vs = 72 km s–1 (C) vs = 62 km s–1 (D) vs = 42 km s–1

1. Solution: (D)

2 e

e

e

GmM1mv

2 R =0 ...(i)

Here ve = 11.2 km/sec

Now 2 e s

s

e

GmM GmM1mv 0

2 R d

5

e2 es 4

e e

Gm 3 10 MGmM1mv

2 R 2.5 10 R

e

e

GmM13.

R

2

e

113. mv

2

s ev v 13 40.38 km/sec

Hence (D) vs = 42 km/s is correct

2. A photoelectric material having work-function 0 is illuminated with light of wavelength

0

hc

. The fastest photoelectron has a de Broglie wavelength d. A change in wavelength

of the incident light by results in a change d in d. Then the ratio d/ is proportional to

(A) d / (B) 3

d / (C) 2 2

d / (D) 3 2

d /

2. Solution: (D)

0 max

hcKE

and d

h

2m KE

Hence d

0

h

hc2m



2

2

d

0

h

hc2m

02 2

d

1 2m hc

h

d3 2 2

d

2 2m hc. .

h

3

d d

2

3. A person measures the depth of a well by measuring the time interval between dropping a stone

and receiving the sound of impact with the bottom of the well. The error in his measurement of

time is T = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the

acceleration due to gravity g = 10 ms–2 and the velocity of sound is 300 ms–1. Then the fractional

error in the measurement, L/L, is closest to

(A) 1% (B) 3% (C) 0.2% (D) 5%

3. Solution: (A)

sound

2L LT

g V

sound

1 2 LT . L

g V2 L

0.01 = 2 L

. L3002 20 10

0.01 = 1 1

L20 300

L = 0.1875

L 0.1875 L

100 1%L 20 L

4. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in

the figure. The distance between the diametrically opposite vertices of the star is 4a. The

magnitude of the magnetic field at the center of the loop is

(A) 0I 6 3 14 a

(B) 0I 3 3 14 a

(C) 0I 3 2 34 a

(D) 0I 6 3 14 a



4. Solution: (D)

d = 2a sin30o = a

CB due to each segment 0I 3 1

4 a 2 2

0I 3 18 a

There are 12 segments

Hence 0IB . 6 3 64 a

5. Consider 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. The

rate of fractional change in density1 d

dt

is constant. The velocity v of any point on the surface

of the expanding sphere is proportional to

(A) 1

R (B) R3 (C) R (D) R2/3

5. Solution: (C)

4

M . R3

M 4

R3

2

2

M d 4 dR. .3R .

dt 3 dt

Hence 2

dR 1R

dt R

6. 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 horizontal surface

about the leading vertex without slipping and sliding as depicted. The maximum increase in

height of the locus of the center of mass for each polygon is . Then depends on n and h as

(A) 2h sinn

(B) 2h tan2n

(C) 1

h

cos 1n

(D) 2

h sinn

6. Solution: (C)

Hence h sec hn

h sec 1n

1h 1

cosn



7. 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) 2S 1 b P bQ (B) S 1 b P bQ

(C) 2S 1 b P b Q (D) S b 1 P bQ

7. Solution: (B)

Q P R

S P bR P b Q P

S P 1 b bQ

SECTION 2 (Maximum Marks: 28)


• 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.


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.



• For example, if (A), (C) and (D) are all the correct options for a question, darkening all these

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.

8. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as

shown in the figure. Which of the following statements is/are correct

(A) The electric flux passing through the curved surface of the hemisphere is 0

Q 11

2 2

(B) Total flux through the curved and the flat surface is 0

Q

(C) The component of the electric field normal to the flat surface is constant over the surface

(D) The circumference of the flat surface is an equipotential



8. Solution: (A, D)

2 2

0

1 QE .

4 R r

d = E.ds 2 2 2 2

0

Q R. .2 r dr

4 R r R r

R

3/20 2 20

QR r drd

2 R r

R

2 20 0

QR 1

2 R r

0

QR 1 1

2 R2 R

0

Q 11

2 2

Hence curved = – 0

Q 11

2 2

Total flux will be zero

The circumference of flat surface will be equipotential

9. 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 bar with the vertical is

. Which of the following statements about is motion is/are correct?

(A) When the bar makes an angle with the vertical, the displacement of its midpoint from the

initial position is proportional to (1 – cos)

(B) Instantaneous torque about the point in contact with the floor is proportional to sin

(C) The trajectory of the point A is a parabola

(D) The midpoint of the bar will fall vertically downward

9. Solution: (A, D)

L L L

y cos 1 cos2 2 2

Let coordinates of A(x, y)

L

x sin2

L L

y cos2 2

Hence path will be circular.

Instantaneous Torque due to mg and pseudo force will not be proportional to sin.



10. A uniform magnetic field B exists in the region between x = 0 and x = 3R

2 (region 2 in the

figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum

p directed along x-axis enters region 2 from region 1 at point P1 (y = – R).

(A) For B > 2 p

3 QR, the particle will re-enter region 1

(B) When the particle re-enters region 1 through the longest possible path in region 2, the

magnitude of the change in its linear momentum between point P1 and the farthest point

from y-axis is p/ 2

(C) For a fixed B, particles of same charge Q and same velocity v, the distance between the pint

P1 and the point of re-entry into region 1 is inversely proportional to the mass of the particle

(D) For B = 8 p

13 QR, the particle will enter region 3 through the point P2 x-axis

10. Solution: (A, D)

Particle will reach region 1 if r <3R

2

p 3R

2QB 2

2P

B3QR

Change in momentum will be equal to 2P distance between P and pint of sentry

2P

d 2r mQB

2

22 3Rr r R

2

2

2 2 29Rr r R 2rR

4

213R

2rR4

13R P

r8 qB

8P

B13QR



11. 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 continuously

applied on the surface of the wheel so that it just climbs the step

without slipping. Consider the torque about an axis normal to the

plane of the paper passing through the pint Q. Which of the following

option is/are correct

(A) If the force is applied normal to the circumference at point P then is zero

(B) If the force is applied normal to the circumference at point X then is constant

(C) If the force is applied tangentially at point S then 0 but the wheel never climbs the step

(D) If the force is applied at point P tangentially then decreases continuously as the wheel

climbs

11. Solution: (A, B)

Considering force is constant in magnitude as per question.

12. Two coherent monochromatic point source S1 and S2 of wavelength

= 600 m are placed symmetrically on either side of the center of the

circle as shown. The sources are separated by a distance d = 1.8 mm. This

arrangement produces interference fringes visible as alternate bright and

dark spots on the circumference of the circle. The angular separation

between two consecutive bright spots is . Which of the following

options is/are correct

(A) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000

(B) A dark spot will be formed at the point P2

(C) The angular separation between two consecutive bright spots decreases as we more from P1

to P2 along the first quadrant.

(D) At P2 the order of the fringe will be maximum

12. Solution: (A, D)

minx dcos , X 0 at = 90o

maxx d 1.8mm n

3

9

1.8 10n 3000

600 10

At P2 Xmax = n hence BF.

x n dcos

n d sin .

1

sin

if decrease increases.



13. A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2

through a switch S as shown. There is no mutual inductance between the two inductors. The

switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the

following option is/are correct

(A) After a long time, the current through L2 will be 1

1 2

LV

R L L

(B) The ratio of the current 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

LV

R L L

(D) At t = 0, the current through the resistance R is V/R

13. Solution: (A, B, C)

Itotal = V

R and it will be divided inversely proportional to L.

Hence 1 22 1

1 2 1 2

L LV Vi . , i .

R L L R L L

1 2

2 1

i L

i L constant.

At t = 0, i = 0

14. The instantaneous voltages at three terminals marked X, Y and Z are given by

X 0V V sin t

Y 0

2V V sin t

3

and

Z 0

4V V sin t

3

.

An ideal voltmeter is configured to read rms value of the potential difference between its

terminals. It is connected between points X and Y and then between Y and Z. The reading (s) of

the voltmeter will be

(A) rms

XY 0

1V V

2

(B) independent of the choice of the two terminals

(C) rms

XY 0V V

(D) rms

XY 0

3V V

2

14. Solution: (B, D)

VXY = V0sint – V0sin2

t3

.

0 0

2V sin t V sin t

3



rms

0

XY 0

V 3 3V V

22

YZ 0 0

4V V sin t V sin t

3 3

0 0

4V sin t V sin t

3 3

rms

0

YZ

V 3V

2


• This section contains TWO paragraphs.

• Based on each paragraph, there are TWO questions.


correct.




darkened.

Zero Marks : 0 In all other cases

PARAGRAPH – 1

Consider a simple RC circuit as shown in figure 1.

Process 1: In the circuit the switch S is closed at t = 0 and the capacitor is fully charge to voltage

V0 (i.e. charging continues for time T >> RC). In the process some dissipation (ED) occurs across

the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.

Process 2: In a different process the voltage is first set to 0v

3 and maintained for a charging time

T>> RC. Then the voltage is raised to 02v

3 without discharging the capacitor and again

maintained for a time T >> RC. The process is repeated one more time by raising the voltage to

V0 and the capacitor is charged to the same final voltage V0 as in Process 1.

These two processes are depicted in Figure 2.

15. In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are

related by:

(A) EC = 2ED (B) EC = ED ln 2 (C) EC = ED (D) EC = 1

2 ED



15. Solution: (C)

During charging, EC = ED = 21

CV2

16. In Process 2, total energy dissipated across the resistance ED is :

(A) 2

D 0E 3CV (B) 2

D 0

1E 3 CV

2

(C) 2

D 0

1 1E CV

3 2

(D) 2

D 0

1E CV

2

16. Solution: (C)

From t = 0 to t = T

1

2

20D 0

V1 1 1E C CV

2 3 2 9

From t = T to t = 2T.

2

2

20D 0

V1 1 1E C CV

2 3 2 9

From t = 2T to t =

3

2

D 0

1 1E CV

2 9

Hence 1 2 3D D D DE E E E 2

0

1 1CV

3 2

PARAGRAPH – 2

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in

Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger

traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by

the point where the ring and the finger is in contact is r. The finger rotates with an angular

velocity 0. The rotating ring rolls without on the outside of a smaller circle described by the

point where the ring and the finger is in contact (Figure 2). The coefficient of friction between

the ring and the finger is and the acceleration due to gravity is g.

17. The total kinetic energy of the ring is

(A) 2 2

0M R (B) 22

0M R r (C) 22

0

3M R r

2 (D)

22

0

1M R r

2

17. Solution: (BONUS)

Centre of ring moves in a circular path of radius (R–r) with same angular velocity o. This is

motion of COM of ring and can be termed as Translatory Motion hence.

KECOM = 2

CM

1MV

2 =

2 2

o

1M R r

2

As there is no slipping between finger and surface of ring hence angular speed of ring with

respect to its COM is related with o as

R = ro



hence = or

R

hence, KEring, COM = 2 21MR '

2 = 2 2

o

1M r

2

KEtotal = KEring, COM + KECOM

= 22 2

o

1M R r r

2

18. The minimum value of 0 below which the ring will drop down is

(A) 3g

2 R r (B)

g

2 R r (C)

2g

R r (D)

g

R r

18. Solution: (D)

2

oM R r Mg

o

g

(R r)

Hence, (D) is correct answer.

END OF PART I : PHYSICS



PART II : CHEMISTRY




correct.




darkened.



19. 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 of water as 2 K kg

mol–1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature

(T). [molecular weight of ethanol is 46 g mol–1]

Among the following, the option representing change in the freezing point is

(A) (B)

(C) (D

19. Solution: (D)

Tf = Kf m

Tf = 2 × 34.5 1000

46 500

Tf = 3

So freezing point of solution = 270 K

20. The standard state Gibbs free energies of formation of C(Graphite) and C(diamond) at T = 298

K are

fG°[C(graphite)] = 0 kJ mol–1

fG°[C(diamond)] = 2.9 kJ mol–1.

The standard state means that the pressure should be 1 bar, and substance should be pure at a

given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces



its volume by 2 × 10–6 m3 mol–1. If C(graphite) is converted to C(diamond) isothermally at

T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is

[Useful information : 1 J = 1 kg m2 s–2 ; 1 kg m–1 s–2 ; 1 bar = 105 Pa]

(A) 14501 bar (B) 1450 bar (C) 58001 bar (D) 29001 bar 20. Solution : (A)

Cgraphite Cdiamond

G° = 2.9 KJ/mole

P × 2× 10–6 = 2.9 × 103

P = 92.910

2 Pa = 42.9

102 bar

P = 14500 bar

P2 = 14501 bar

21. For the following cell,

Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)

when the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for G

(in J mol–1) is

[F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V]

(A) 2.303 RT + 1.1 F (B) 2.303 RT – 2.2F

(C) –2.2 F (D) 1.1F

21. Solution : (B)

Zn + Cu2+ Zn2+ + Cu(s)

Ecell = cellE – 2

2

0.0591 | Zn |log

2 | Cu |

G = – nF Ecell

G

2F

= 1.1 –

0.0591

2 log 10

G = – 2.2F + 0.0591 F

G = 2.303 RT – 2.2 F

22. The major product of the following reaction is

2(i) NaNO , HCl,0 C

(ii) aq. NaOH

(A) (B)

(C) (D)



22. Solution: (B)

23. The order of basicity among the following compounds is

(I) (II) (III) (IV)

(A) II > I > IV > III (B) IV > I > II > III

(C) IV > II > III > I (D) I > IV > III > I

23. Solution: (B)

2

2

NH|

NH C NH is most basic as out of 3, 1 nitrogen is always ready to donate lone pair and III

is aromatic so least basic

24. Which of the following combination will produce H2 gas?

(A) Au metal and NaCN(aq) in the presence of air

(B) Fe metal and conc. HNO3

(C) Zn metal and NaOH(aq)

(D) Cu metal and conc. HNO3

24. Solution: (C)

Zn + 2NaOH — Na2ZnO2 + H2

25. The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3, and H4P2O6

is

(A) H3PO4 > H4P2O6 > H3PO3 > H3PO2 (B) H3PO4 > H3PO2 > H3PO3 > H4P2O6

(C) H3PO3 > H3PO2 > H3PO4 > H4P2O6 (D) H3PO2 > H3PO3 > H4P2O6 > H3PO4

25. Solution: (A)

Oxidation no. of P in H3PO2 + 1, H3PO4 + 5, H3PO3 + 3 & H4P2O6 + 4


















26. For a reaction taking place in a container in equilibrium with its surroundings, the effect of

temperature on its equilibrium constant K in terms of change in entropy is described by

(A) With increase in temperature, the value of K for exothermic reaction decreases because the

entropy change of the system is positive.

(B) With increase in temperature, the value of K for endothermic reaction increases because

the entropy change of the system is negative.

(C) With increase in temperature, the value of K for exothermic reaction decreases because

favourable change in entropy of the surroundings decreases.

(D) With increase in temperature, the value of K for endothermic reaction increases because

unfavourable change in entropy of the surroundings decreases.

26. Solution: (D)

G° = H° – TS°

for exothermic reaction H < 0 & on increasing temp. equilibrium constant decreases.

for endothermic reaction H > 0 & on increasing temp. equilibrium constant increases.

27. Among the following, the correct statement(s) is(are)

(A) Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure.

(B) AlCl3 has the three-centre two-electron bonds in its dimeric structure.

(C) BH3 has the three-centre two-electron bonds in its dimeric structure.

(D) The Lewis acidity of BCl3 is greater than that of AlCl3 27. Solution: (A, C)

AlCl3 In dimer form has 3C – 4e– bond BCl3 has less Lewis acidity due to back bonding.

28. The correct statement(s) about surface properties is(are)

(A) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system.

(B) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The

adsorption of ethane will be more than that of nitrogen on same amount of activated

charcoal at a given temperature.

(C) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is

dispersion medium.

(D) Brownian motion of colloidal particles does not depend on the size of the particles but

depends on viscosity of the solution.




For Adsorption H < 0 < & S < 0

Easily liquefiable gases will be more adsorbed on charcoal. Cloud is aerosol (Liquid is Gas

colloidal solution).

Brownian motion is independent of nature of colloid but depends on size of particle and viscosity

of the solution.

29. For the following compounds, the correct statement(s) with respect to nucleophilic substitution

reactions is (are)

(I) (II) (III)

3

3

3

CH|

H C C Br|

CH

(IV)

(A) I and III follow SN1 mechanism

(B) I and II follow SN2 mechanism

(C) Compound IV undergoes inversions of configuration.

(D) The order of reactivity for I, III and IV is : IV > I > III 29. Solution: (A, B, C, D)

can give SN1 as well as SN

2 mechanism

gives SN2 mechanism

3

3

3(III)

CH|

|CH

CH C Br gives SN1 mechanism

can give retention as well as invertion

Order of reactivity IV > I > III

30. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The

correct option(s) among the following is(are)

(A) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used.

(B) The activation energy of the reaction is unaffected by the value of the steric factor

(C) Experimentally determined value of frequency factor is higher than that predicted by

Arrhenius equation.

(D) The value of frequency factor predicted by Arrhenius equation is higher than that

determined experimentally.

30. Solution: (B, C)

Static Factor P = experimental

Theoretical

A

A



31. The option(s) with only amphoteric oxides is (are)

(A) Cr2O3, BeO, SnO, SnO2 (B) ZnO, Al2O3, PbO, PbO2

(C) Cr2O, CrO, SnO, PbO (D) NO, B2O3, PbO, SnO2


CrO is basic and NO is neutral oxide.

32. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of

Q and S is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S

undergoes haloform reaction but not Cannizzaro reaction.

(i) P 3 2 2

28 8

(i) O /CH Cl

(ii) Zn/H O(C H O)

Q (ii) R 3 2 2

28 8

(i) O /CH Cl

(ii) Zn/H O(C H O)

S

The option(s) with suitable combination of P and R, respectively, is (are)

(A)

(B)

(C)

(D)


Q is

or S is







correct.




darkened.


PARAGRAPH - 1

Upon heating KClO3 in the presence of catalytic amount of MnO2, a gas W is formed. Excess

amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives

Y and Z.

33. Y and Z are, respectively

(A) N2O3 H3PO4 (B) N2O5 and HPO3 (C) N2O4 and HPO3 (D) N2O4 and H3PO3

33. Solution: (B)

HNO3 + P4O10 — HPO3 + N2O5

N2O5 is anhydride of HNO3.

34 W and X are, respectively

(A) O2 and P4O10 (B) O2 and P4O6 (C) O3 and P4O6 (D) O3 and P4O10 34. Solution: (A)

2KClO3 2KCl + 3O2

P4 + O2 (excess) — P4O10

PARAGRAPH - 2

The reaction of compound P with CH3MgBr (excess) in (C2H5)2 followed by addition of H2O

gives Q. The compound Q on treatment with H2SO4 at 0°C gives R. The reaction of R with

CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O

produces compound S. [Et in compound P is ethyl group]

—— Q —— R —— S

35. The reactions, Q to R and R to S, are

(A) Dehydration and Friedel-Crafts acylation

(B) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation

(C) Aromatic sulfonation and Friedel-Crafts acylation.

(D) Friedel-Crafts alkylation and Friedel-Crafts acylation.



36. The product S is

(A) (B)

(C) (D)

35. Solution: (A) 36. Solution: (B)

END OF PART II : CHEMISTRY



PART III : MATHEMATICS




correct.




darkened.



37. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes

2x + y – 2z = 5 and 3x – 6y – 2z = 7, is

(A) 14x + 2y + 15z = 31 (B) 14x – 2y + 15z = 27

(C) 14x + 2y – 15z = 1 (D) –14x + 2y + 15z = 3

37. Solution: (A)

x 1 y 1 z 1

2 1 2 0

3 6 2

14x 2y 15z 31

38. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OP.OQ OR.OS OR.OP OQ.OS OQ.OR OP.OS

Then the triangle PQR has S as its

(A) Centroid (B) Circumcentre (C) Orthocentre (D) Incenter

38. Solution: (C)

p.q r.s r.p q.s p s . q r 0 PS QR

Similarly QS PR and RS PQ S is orthocentre of PQR

39. Let S = {1, 2, 3,…..,9}. For k = 1, 2,……, 5, let Nk be the number of subsets of S, each containing

five elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 =

(A) 252 (B) 125 (C) 210 (D) 126

39. Solution: (D) N1 + N2 + N3 + N4 + N5 = [Total number of subsets which contain exactly 5 elements] – [No]

= 9C5– 0 = 126

40. Three randomly chosen non-negative integers x, y and z are found to satisfy the equation

x + y + z = 10. Then the probability that z is even, is

(A) 1

2 (B)

6

11 (C)

36

55 (D)

5

11

40. Solution: (B)

x + y + z = 10, x, y, z 0

Total number of solutions = 10 2 12

2 2C C 66

No. of solutions when z is even = z 0 z 10z 2 z 4 z 6 z 8

11 9 7 5 3 1 36

Required probability = 36 6

66 11



41. How many 3 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the

diagonal entries of MTM is 5?

(A) 135 (B) 126 (C) 162 (D) 198

41. Solution: (D)

Let M =

a p x

b q y

c r z

tr (MTM) = a2 + b2 + c2 + p2 + q2 + r2 + x2 + y2 + z2 = 5

Total no. of ways = 9P2 + 9C5 = 72 + 126 = 198, where

9P2 = No. of matrices when 2 comes once, 1 comes once and 0 comes 7 times.

9C5 = No. of matrices when 1 comes 5 times and 0 comes 4 times.

42. If y = y(x) satisfies the differential equation

1

8 x 9 x dy 4 9 x dx, x 0

And y(0) = 7 , then y(256) =

(A) 16 (B) 9 (C) 3 (D) 80

42. Solution: (C)

dx

dy

8 x 9 x 4 9 x

y 256

07

dxdy

8 x 9 x 4 9 x

256

0

y 7 4 9 x 3 7

y = 3

43. If f : is a twice differentiable function such that f (x) > 0 for all x , and

1 1f , f (1) 1

2 2

, then

(A) f (1) > 1 (B) f (1) 0 (C) 0 < f (1) 1

2 (D)

1

2< f (1) 1

43. Solution: (A)

f (x) > 0 and f1 1

2 2

, f(1) = 1

Applying LMVT on 1

,12

we get

1f (1) f

2f '(c) 1

11

2

for some 1

c ,12

Also f (x) is a strictly increasing function (f (x) > 0)

f (1) > 1


















44. Let and be nonzero real numbers such that 2 (cos – cos ) + cos cos . Then which

of the following is/are true?

(A) 3 tan tan 02 2

(B) tan 3 tan 0

2 2

(C) 3 tan tan 02 2

(D) tan 3 tan 0

2 2

44. Solution: (A, C)

2(cos – cos) + cos cos = 1

2cos2 2

cos2

2

+ cos2

2

– 3 cos2

2

= 0 [use cos 2 = 2 cos2 –1]

2 = 3 sec2

2

– sec2

2

3 tan2 2

= tan2

2

[use sec2 = 1 + tan2 ]

45. If g(x)sin(2x)

1

sin xsin (t)dt , then

(A) g ' 22

(B) g ' 2

2

(C) g ' 22

(D) g ' 22

45. Solution: (BONUS)

g(x) = 2 cos 2x sin–1(sin 2x) – cos x sin–1(sin x)

at x = 2

, g(x) = 2( – 2x) cos 2x – x cos x g ' 0

2

at x = 2

, g(x) = –2( + 2x) cos 2x – x cos x g ' 0

2

46. If f(x) =

cos(2 x) cos(2 x) sin(2 x)

cos x cos x sin x

sin x sin x cos x

, then

(A) f(x) attains its maximum at x = 0

(B) f (x) = 0 at exactly three points in (–, )

(C) f (x) = 0 at more than three points in (–, )

(D) f (x) attains its minimum at x = 0



46. Solution: (A, C)

f(x) =

cos 2x cos 2x sin 2x

cos x cos x sin x cos 2x cos 4x

sin x sin x cos x

f(x) attains its maximum when cos 2x = 1 i.e., x = 0

f (x) = – 2 sin 2x (1 + 4 cos 2x) = 0 at exactly 7 points in (–, )

47. Let f(x) =1 x(1 |1 x |) 1

cos|1 x | 1 x

for x 1. Then

(A) x 1

lim f (x) 0 (B)

x 1lim f (x)

does not exist

(C) x 1

lim f (x)= 0 (D)

x 1lim f (x)

does not exist

47. Solution: (AD)

f(x) = 1 x(1 |1 x |) 1

cos|1 x | 1 x

x 1 x 1 x 1

1 x(1 1 x) 1 1lim f (x) lim cos lim 1 x cos 0

1 x 1 x 1 x

x 1 x 1 x 1

1 x(1 x 1) 1 1lim f (x) lim cos lim x 1 cos

x 1 1 x 1 x

does not exist

48. If the line x = divides the area of region R = {(x, y) 2 : x3 y x, 0 x 1} into two equal

parts, then

(A) 1

12 (B) 4 22 4 1 0 (C)

10

2 (D) 4 24 1 0


1

3 3

0 0

2 (x x )dx (x x )dx

2 4 1 1 1

22 4 2 4 4

24 – 42 + 1 = 0

A root of above equation lies in 1

,12

[apply IVT on 2x4 – 4x2 + 1 in 1

,12

]

49. If f : is a differentiable function such that f (x) > 2f(x) for all x , and f(0) = 1, then

(A) f(x) > e2x in (0, ) (B) f(x) is increasing in (0, )

(C) f (x) < e2x in (0, ) (D) f(x) is decreasing in (0, ) 49. Solution: (A, B)

f (x) > 2f(x)

e–2x f (x) – 2e–2x f(x) > 0

2xe . f (x) ' 0

g(x) = e–2x f(x) is an increasing function

x > 0 g(x) > g(0) e–2x f(x) > 1 f(x) > e2x

Now f(x) > e2x x (0, ) f(x) > 0 x (0, )

f (x) > 2f(x) f (x) > 0 x (0, )

f is increasing on (0, )



50. If k 198

k 1 k

k 1I dx

x(x 1)

, then

(A) I > loge 99 (B) I < loge 99 (C) I < 49

50 (D) I >

49

50

50. Solution: (B, D)

x + 1 > k +1 [As k < x < k +1]

1 1

x 1 k 1

1 1

x(x 1) x(k 1)

k 1 k 1

k k

dx dx

x(x 1) x(k 1)

k 1 k 198 98

k 1 k 1k k

(k 1)dx dx

x(x 1) x

k 198 98

k 1 k 1k

(k 1)dx k 1ln

x(x 1) k

I < ln(99)

Also, k 1 k 1 1

orx(x 1) (k 1)100 100

[As 1 k 98]

k 1 k 1

k k

k 1 dxdx

x(x 1) 100

98

k 1

kI

100

49

I50





correct.




darkened.


PARAGRAPH - 1

Let O be the origin, and OX , OY and O Zbe three unit vectors in the directions of the sides

QR , RP , PQ , respectively, of a triangle PQR.

51. | OXOY | =

(A) sin 2R (B) sin (P + R) (C) sin (P + Q) (D) sin (Q + R)



51. Solution: (C)

OX OY OX . OY sin XOY

= sin R

= sin (P + Q)

52. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos (Q + R) + cos (R + P) is

(A) 5

3 (B)

5

3 (C)

3

2 (D)

3

2

52. Solution: (D)

cos (P + Q) + cos (Q + R) + cos (R + P)

= –(cos P + cos Q + cos R)

3

2

,

Let x = – (cos P + cos Q + cos R)

–x = 2P Q P Q R

2cos cos 1 2sin2 2 2

2 R

2sin2

– P Q R

2cos sin (1 x) 02 2

D 0 2 P Q4cos 8(1 x) 0

2

2 P Q

2x 2cos 1 32

3

x2

PARAGRAPH - 2

Let p, q be integers and let , be the roots of the equation, x2 – x – 1 = 0, where . For

n = 0, 1, 2, ….., let an = pn + qn.

Fact : If a and b are rational numbers and a + b 5 = 0, then a = 0 = b.

53. a12 =

(A) 2a11 + a10 (B) a11 + a10 (C) a11 + 2a10 (D) a11 – a10

53. Solution: (B)

2x x 1 0 ,

2 1 0 and 2 1 0

12 11 10 0 and 12 11 10 0

12 11 10 12 11 10p q 0

12 11 10a a a 0

54. If a4 = 28, then p + 2q =

(A) 12 (B) 7 (C) 14 (D) 21



54. Solution: (A)

2 1 0 2 1 4 2 2 1

4 3 2

a4 = 28

p(3 + 2) + q (3 + 2) = 28

7 3 5 7 3 5

p q 282 2

7 3 5

p q p q 282 2

p – q = 0 and p + q = 8

p = 4, q = 4

p + 2q = 12

END OF PART III : MATHEMATICS

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