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21 May 2017 JEE ADVANCED 2017 P1 QUESTION PAPER (CODE-3) 3

PART I : PHYSICSSECTION 1(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) 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 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

01. A flat plate is moving normal to its plane through a gas the action of a constant force F. The gasis kept at a very low pressure. The speed of the plate v is much less than the average speed uof the gas molecules. Which of the following options is/are true?A) The resistive force experienced by the plate is proportional to vB) At a later time the external force F balances the resistive forceC) The plate will continue to move with constant non-zero acceleration, at all timesD) The pressure difference between the leading and trailing faces of the plate propertional to uvKey : B, CSol : A) Assume the plate is at rest. Let n be the no.of molecules per unit volume

A be the Area of plate m be the mass of each molecule

p of one molecule = 2 mV (being elastic collism)

No.of molecules striking per unit time= nAV

The corresponding resistive force 2f nAV mU , 22f nAmV

2f V

B) 2 2

1 1log

2

a x

a x a a x

Fdv

F f m isdt

of the form

2 dVF KV m

dt

or 2

1 1dv dt

F KV m

Integrating ' 'V is of the form

1

1

kt

kt

eV C

e

As t V C

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JEE Advanced Paper -1 Question Paper with Solutions and Key


Therefore at a later time F balances fC) At all times the acceleration of the plate is non-zero

D) Pressure on rear side of plate is proportional to 2uPressure on front side of plate has two components

(1) pressure proportional to 2u (2) pressure propotional to 2v

Net pressure is propartional to 2v

02. A human body has a surface area of approaximately 1 2 .m The normal body temperature is 10

K above teh surrounding room temperature 0T . Take the room temperature to be 0 300 .T K

For 0 300 ,T K the value of 3 20 460T Wm (Where is the stefan Boltzmann constant).

Which of the following options is/are correct?

A) If the surrounding temperature reduces by a small amount 0 0 ,T T then to maintain the same body

temperature the same (living) human being needs to radiate 30 04W T T more energy per unit time

B) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiationemitted by the body would shift to longer wavelengthsC) Reducing teh exposed surface area of the body (e.g. by curling up allows humans to maintain the samebody temperature while reducing teh energy lost by radiationD) The amount of energy radiated by the body in 1 second is close to 60 JoulesKey : A,C,D

Sol : A] 440[ ]W A T T

differentiating

30 04W A T T asTis to beconstant

as 0T is - re (reducing)

30 04W A T T

3 20 04 1W T T A m

as W is positive, he has to radiate more energy per unit time.03. A block M hangs vertically at teh bottom end of a uniform rope of constant mass per unit length.

The top end of the rope is attached to fixed rigid support at O. A transverse wave pulse (Pulse 1)

of wavelength 0 is produced at point O on the rope. The pulse takes time 0 AT to reach point A.A.

If the wave pulse of wavelength 0 to reach point A (Pulse 2) without disturbing the position of

M it takes time 0AT to reach point O. Which of the following options is/are correct?SmartPrep.in



A) The time 0 0A AT TB) The velocity of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of ropeC) The wavelength of Pulse 1 becomes longer when it reaches point AD) The velocity of any pulse along the rope is independent of its frequency and wavelengthKey : A,B,D

Sol : Rope is uniform 0 0A AT T Af mid point velocities of two pulses are equal When pulse (1) reaches to bottom point A velocity decreases (tension decreases)

wave length decreases

D) Velocity at any point V xg i.e independent of frequency and wavelength

04. For an isosceles prism of angle A and refractive index , it is found that the angle of minimum

deviation .m A Which of the following options is/are correct?

A) At minimum deviation, the incident angle 1i and the refracting angle 1r at the first refracting surface are

related by 1 1 / 2r i

B) For this prism, the refractive index and the of prism A are related as11

cos2 2

A

C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of

incidence at the first surface is1 2

1 sin sin 4cos 1 cos2

Ai A A

D) For the angle of incidence 1 ,i A the ray inside the prism is parallel to the base of the prism

Key : A C DSol :A) In min div position / 2r A

22 2

A D A Ai A r

B) sin / 2 sin

, 2cos / 2sin / 2 sin / 2

A D AA

A A

1/ 2 cos2

A

12 cos2

A

C) At first face 1sin sinc A C

sin cos cos sinA C A C

2 1 1sin cosA A

2sin , 2 sin 1 cosA A

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2sin 4cos 1 cos2

AA A

D) 122 2

A D A Ai A r i

05. In the circuit shown, 1 , 1 1 .L H C F and R k They are connected in series with an a.c.

source 0V V sin t as shown. Which of the following options is/are correct?

A) At 6 110 .5 ,rad the circuit behaves like a capacitor

B) At 0 the current flowing through the circuit becomes nearly zero

C) The current will be in phase with the voltage if 4 110 .rad s D) The frequency at which the current will be in phase with teh voltage is independent of RKey : B,D

Sol :6

0

1) 10 / secA w rad

LC

for 6. 10w inductive

B)

I

0I

w0w

C) for 40 10w rad/sec, a.c. not in phase with a.v

D) Conceptual06. A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the

borizontal frictionless surface of a fixed table. Initially the right edge of the block is at x=0, in aco-ordinate system fixed to the table, A point mass m is released from rest at the topmost pointof the path as shown and it slides down. When the mass loses contact with the block, its positionis x and the velocity is v. At thet instant, which of the following options is/are correct?

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A) The velocity of the block M is: 2m

V gRM

B) The x component of displacement of the center of mass of the block M is:mR

M m

C) The position of the point mass is: 2mR

xM m

D) The velocity of the point mass m is:

2

1

gRv

m

M

Key : B,D

Sol : w.r.t ground frame1 ( )

. . . .

V vel of block M

V vel vel of m w r t M

1 1MV m V V

1 __________ 1mV

VM m

using LCE,

2 21 11 1________ 2

2 2M V m V V mgR

Solving (1) & (2)

2M m

V gRM

and1 2

mV m M mV gR

M m M m M

2

1 2m gRV

M m M

A)

B) The displacement of block M=mR

M m

C) M x m R x . . .x displ of block M w r t Ground

mRx

M m

displ ofmR MR

m R x RM m M m

D) vel. of point mass m is V-V

mVV

M m

2MV M M m

gRM m M m M

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Vel. of point mass m =

2

1

gRm

M

07. A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown inthe figure. At the point of crossing the wires remains electrically insulated from each other. Theentire loop lies in the plane (of the paper). A uniform magnetic field B

points into the plane ofthe paper. At t=0, the loop starts rotating about the common diameter as axis with a constantangular velocity in the magnetic field. Which of the following options is/are correct?

A) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude ofmaximum emf induced in the smaller loop aloneB) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of thepaperC) The emf induced in the loop is proportional to the sum of the areas of the two loopsD) The net emf induced due to both the loops is proportional to cos tKey : A,B

Sol : 1 0cos , n

de BA t BAW CO Wt e

dt

2 2 0e B AWS inwt e wt

1mgxe BAW

max 22e BAW

Option A closed

sinnete BA t Hence Option [C] correct A & B are correct

SECTION 2 (Maximum Mars: 15) This section contains FIVE questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive For each question, darken the bubble corresponding to the correct integer in the ORS For eavch question, marks will be awarded in one of the following categories:

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Full marks : +3 If only the bubble corresponding to the correct answer is darkenedZero Marks : 0 In all other cases

08. A stationary source emits sound of frequency 0 492 .f Hz The sound is reflected by a large car

approaching the source with a speed of 2 1.ms The reflected signal is received by the source

and superposed with the original. What will be the beat frequency of the resulting signal in ?Hz(Given that the speed of sound in air is 330 1ms and the car reflects the sound at the frequencyit has received).Key : 6

Sol : 492 Zn H

| V un n

V u

| 330 2492 498

330 2n Hz

beats | 6n n n Hz

09. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number in to

another with quantum number .f i fn V and V are respectively the initial and final potential energies

of the electron. If 6.25,i

f

V

V then the smallest possible fn is

Key : 5Sol : PE = 2 TE

2

13.6 2i

i

Vn

2

13.6 2f

f

Vn

2

2

625

100fi

f i

nV

V n

25 5

10 2f

i

n

n

Min value of 5fn

10. A drop of liquid of radius 210R m having surface tension10.3

4S Nm

divides itself into K

identical drops. In this process the total change in the surface energy 310U . If 10K then the value of isKey : 6

Sol : 2 2 / 34U R n n

3 4 2 / 30.110 4 10

4x x

Soving = 610k

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11. 3 1 is an isotope of iodine that decays to an isotope of Xenon with a half - life of 8 days. AA

small amount of a serum labelled with 3 1 is injected into the blood of a person. The activity of

the amount of 3 1 injected was 524 10 Beequerel (Bq.). It is known that the injected serum willget distributed unifomly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml ofblood is drawn from the person’s body, and gives an activity of 11.5 Bq. The total volume ofblood is drawn from the person’s body, and gives an activity of (you may use

1 1and in 2 0.7xe x for x )

Key : 5

Sol : For activity of 52.4 10 Bd volume of blood

532.4 10

2.5 10120.76

12. A monochromatic light is travelling in a medium of refractive index 1.6 . It enters a stack of

glass layers from the bottom side at an angle 030 . The interfaces of the glass layers areparallel to each other. The refractive indices of different glass layers are monotonically decreasing

as mn n m n , where mn is the refractive index of the thm slab and 0.1n (sec the figure).

The ray is refracted out parallel to the interface between the 1th

m and thm slabs from the

right side of the stack. What is the value of m?Key : 8

Sol : I intiger2

1

sin

sin

i

r

1.6 sin 30 1.6 1 1.6sin

1.5 sin 1.5 2 3r

r

Second intiger 12 1sin sinr r

1 11.6 1.6 1 1.61.5 1.4sin sin

3 2 1.4 2.8r r

III intiger 2 1 111 sinRr r

111.61.4 1.6sin

2.8r

11 1.6sin

2.6r

3

1.6sin

2.4IV r

4

1.6sin

2.2V r

5

1.6sin

2.0VI r

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A c h a r g e d p a r t i c le ( e le c t r o n o r p r o t o n ) i s i n t r o d u c e d a t t h e o r i g i n

0 , 0 , 0x y z w i t h a g iv e n i n i t i a l v e l o c i t y v

A u n i f o r m e l e c t r ic f i e l d E

a n d a

u n i f o r m m a g n e t i c f i e l d B

e x i s t e v e r y w h e r e , T h e v e l o c i t y v

, e l e c t r i c f ie l d E

a n dm a g n e t i c f i e l d B

a r e g iv e n i n c o l u m n s 1 ,2 a n d 3 , r e s p e c t iv e l y , T h e q u a n t i t i e s 0 0,E B

a r e p o s it i v e i n m a g n i t u d e .C o l u m n 1 C o l u m n 2 C o l u m n 3

( I ) E l e c t r o n w it h

0

0

2E

v xB

( i ) 0E E z

( P ) 0B B x

( I I ) E l e c t r o n w it h

0

0

Ev y

B

( i i ) 0E E y

( Q )

0B B x

( I I I ) P r o t o n w i t h 0v

( i i i ) 0E E x

( R )

0B B y

( I V ) P r o t o n w i t h

0

0

2E

v xB

( i v )

0E E x

( S ) 0B B z

SECTION 3 (Maximum Marks : 18) This Section contains SIX questions of matching type This section contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE questions Each questions 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 in darkenedNegative Marks: -1 In all other cases

13. In which case would the particle move in a straight line along the negative direction of y axis

(i.e., move along - y ) ?

A) (III) (ii) (P) B) (II) (iii) (Q) C) (IV) (ii) (S) D) (III) (ii) (R)Key : ASol : The resultant face must be along -y direction i.e acording to the condition of ‘A’. It is possible.

14. In which ease will the particle descrive a helical path with axis along the positive z direction?A) (II) (ii) (R) B) (III) (iii) (Q) C) (IV) (ii) (R) D) (IV) (i) (S)Key : DSol : For helical path E,B direction must be perpendicular to V

15. In which case will the particle move in a straight line with constant velocity ?A) (II) (iii) (S) B) (III) (ii) (R) C) (IV) (i) (S) D) (III) (iii) (P)Key : DSol : Resultant force acting on the charged particle must be zero.

16. Which one of the following options correctly represents a thermodynamic process that is used asa correction in the determination of the speed of sound in an ideal gas?A) (1) (ii) (Q) B) (I) (iv) (Q) C) (IV) (ii) (R) D) (III) (iv) (R)Key : BSol : According to the given table let us first range accadinngly such that

workd are in adiabatic moces. 2 2 1 11

1w p v p v

r

i.e (I) ------- (IV)

workdone in isobaric process 2 1w pv pv i.e II -------- ii

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wark done in isothermal2

1

logv

w ntv

i.e IV -- --------- i

Adiabatic candition for sound propagation

17. Which one of the following options is the correct combination?A) (II) (iv) (P) B) (III) (ii) (S) C) (II) (iv) (R) D) (IV) (ii) (S)Key : BSol : workdone Isobanic mocen

18. Which of the following options is the only correct representation of a process in which

U Q P V ?A) (III) (iii) (P) B) (II) (iii) (P) C) (II) (iii) (S) D) (II) (iv) (R)Key : BSol : It is the condition for isobanic process.

PART - II CHEMISTRYSECTION 1 (Maximum Marks:15)







19. The IUPAC name(s) of the following compound is(are)

A) 1- chloro -4-methylbenzene B) 1- methyl -4- chlorobenzeneC) 4- chlorotoluene D) 4- methylchlorobenzeneKey : A, CSol : A) 1- chloro -4-methylbenzeneC) 4- chlorotoluene

20. The correct statement(s) about the oxoacids, 4HClO and HClO , is(are)

A) The central atom in both 4HClO and HClO is 3sp hybridized

B) The conjugate base of 4HClO is weaker base than 2H O

C) 4HClO is formed in the reaction between 2Cl and 2H O

D) 4HClO is more acidic than HClO because of the resonance stabilization of its anion

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Key : A, B, D

Sol : (A) Hybridisation is 3sp in 4 &HClO HClO

(B) 44 Strong acid Week BaseHClO ClO

2 Strong BaseWeek AcidH O OH

4ClO is weaker base than 2H O(D) Conceptual

21. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the

mole fraction of M in solution is shown in the following figure. Here Lx and Mx represent molefractions of L and M, respectively, in the solution. The correct statement(s) applicable to thissystem is(are)

A) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when 0Lx

B) The point Z epresents vapour pressure of pure liquid M and Raoult’s law is obeyed from 0Lx to

1Lx C) Attractive intermolecular interactions between L-L in pure liquid L and M- M in pure liquid M arestronger than those between L- M when mixed in solution.

D) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when 1Lx Key : C, DSol : Conceptual

22. Addition of excess aqueous ammonia to a pink coloured aqueous solution of 2 2.6MCl H O (X) and

4NH Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex YYbehaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results inthe formation of a blue coloured complex Z. The calculated spin only magnetic moment of X andZ is 3.87 B.M., whereas it zero for complex Y. Among the following options, which statement(s)is(are) correct?A)Z is a tetrahedral complex

B) The hybridization of the central metal ion in Y is 2 3d sp

C) When X and Z are in equilibrium at 00 C , the colour of the solution is pinkD) Addition of silver nitrate to Y gives only two equivalents of silver chlorideKey : A, B, CSol:Step (1)

42 2 2 2 3 36 6

2 3

.6 aquous excess NH Clair

n

MCl H O M H O Cl M NH Cl

Pink X Y d sp Hyb

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Step(2)

0

2 32 2 46 0

Excess HCl n

CM H O Cl MCl Hyb sp

Pink X Z Blue Tetrahedral

23. The correct statement(s) for the following addition reactions is (are)

A) Bromination proceeds through trans-addition in both the reactionsB) O and P are identical moleculesC) (M and O) and ( N and P) are two pairs of distereomersD) (M and O) and (N and P) are two pairs of enantiomersKey : A, CSol :

(A)(B) (M &O) and (W & P) are two pairs of diastereomers

24. An ideal gas is expanded from 1 1 1, ,p V T to 2 2 2, ,p V T under different conditions. The correct

statement(s) among the following is (are)

A) The work done on the gas is maximum when it is compressed irreversibly from 2 2,p V to 1 1,p V

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against constant pressure 1p

B) The work done by the gas is less when it is expanded reversibly from 1V to 2V under adiabatic

conditions as compared to that when expanded reversibly from 1V to 2V under isothermal conditions

C) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic

D) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with 1 2T T , and (ii)

positive, if it is expanded reversibly under adiabatic conditions with 1 2T TKey : A, B, CSol : Conceptual

25. The colour of the 2X molecules of group 17 elements changes gradually from yellow to violetdown the group. This is due to

A) The physical state of 2X at room temperature changes from gas to solid down the group

B) Decrease in * * gap down the groupC) Decrease in HOMO- LUMO gap down the groupD) Decrease in ionization energy down the groupKey : B, CSol : Conceptual

SECTION 2 (Maximum Mars: 15) This section contains FIVE questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive For each question, darken the bubble corresponding to the correct integer in the ORS For eavch 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

26. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined byusing a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodesis 120 cm with an area of cross section of 21cm . The conductance of this solution was found to be

75 10 S . The pH of the solution is 4. The value of limiting molar conductivity 0m of this weak

monobasic acid in aqeous solution is 2 1 110Z S cm mol . The value of Z is

Key : 6Sol :

Cell constl

A

120

1

1120 cm

.l

K CA

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7 1205 10

1K

7600 10K 56 10K

310C

K

C

5 36 10 10

0.0015C

2210

5C

200

5C

40C

H C 410 0.0015

410

0.0015

1

15

0

C

0

1 40

15

0 15 4 10

0 600

20 6 10

6Z 27. The sum of the number of lone pairs of electrons on each central atom in the following species is

2

6 2 3 3, , ,TeBr BrF SNF and XeF

(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)Key : 6Sol: No.of lone pairs

2

6 1TeBr

2 2BrF

3 0SNF

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3 3XeF

Total 628. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of

400pm. If the density of the substance in the crystal is 38 g cm , then the number of atoms

present in 256 g of the crystal is 2410N . The value of N isKey :

Sol : 30

.

.

Z Md

N G

323 8

48

6 10 4 10

M

M= 76.8

76.8 gr 236 10

256 gr ?23

23 24256 6 1020 10 2 10

76.8

N= 229. Among the following the number of aromatic compound(s) is

Key : 4

Sol:

30. Among2 2 2 2 2 2 2 2, , , , , , ,H He Li Be B C N O and 2F , the number of diamagnetic species is

(Atomic numbers: H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9)Key : 6

Sol: 2 2 2 2, 2 2, , , ,H Li Be C N F all diamagnatic

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SECTION 3 (Maximum Marks : 18) This Section contains SIX questions of matching type This section contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE questions Each questions 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 in darkenedNegative Marks: -1 In all other cases

Answer Q-31, Q-32, Q-33 by appropriately matching the information given in the three coloumns ofthe following table

The wave function, , , ln l m is a mathematical function whose value depends upon spherical polar

coordinates , ,r of the electron and characterized by the quantum numbers , ln l and m . Here r

is the distance from nucleus, is colatitude and is azimuth. In the mathematical functions

given in the Table, Z is atomic number and 0a is Bohr radius

Column 1 Column 2 Column 3(I) 1s orbital

(i) 0

3

2

, ,0

l

Zr

an l m

Ze

a

(P)

(II) 2s orbital (ii) One radial node(Q) Probability density at nucleus

30

1

a

(III) 2 zp orbital(iii) 0

5

2 2, , 0

cosl

Zr

an l m

Zre

a

(R) Probability density is maximum atnucleus

(IV) 23 zd orbital (iv) xy- plane is a nodal plane (S) Energy needed to excite electron

from n=2 state to n=4 state is27

32times the energy needed to exciteelectron from n=2 state to n=6 state

31. For hydrogen atom, the only CORRECT combination isA) (II) (i) (Q) B) (I) (iv) (R) C) (I) (i) (S) D) (I) (i) (P)Key : CSol: Conceptual

32. For He ion, the only INCORRECT combination isA) (I) (iii) (R) B) (II) (ii) (Q) C) (I) (i) (S) D) (I) (i) (R)Key : A

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Sol: Conceptual33. For the given orbital in Column 1, the only CORRECT combination for any hydrogen - like

species isA) (IV) (iv) (R) B) (III) (iii) (P) C) (II) (ii) (P) D) (I) (ii) (S)Key : CSol: Conceptual

Answer Q-34, Q-35, Q-36 by appropriately matching the information given in the three coloumnsof the following table

Columns 1,2 and 3 contain starting materials, reaction conditions, and type ofreactions, respectively

Column 1 Column 2 Column 3

(I) Toluene (i) 2/NaOH Br(P) Condensation

(II) Acetophenone (ii) 2 /Br hv (Q) Carboxylation

(III) Benzaldehyde (iii) 3 32/CH CO O CH COOK (R) Substitution

(IV) Phenol (iv) 2/NaOH CO (S) Haloform

34. The only CORRECT combination that gives two different carboxylic acids isA) (II) (iv) (R) B) (III) (iii) (P) C) (I) (i) (S) D) (IV) (iii) (Q)Key: BSol: Conceptual

35. The only CORRECT combination in which the reaction proceeds through radical mechanism isA) (III) (ii) (P) B) (I) (ii) (R) C) (IV) (i) (Q) D) (II) (iii) (R)Key: BSol: Conceptual

36. For the synthesis of benzoic acid, the only CORRECT combination isA) (I) (iv) (Q) B) (III) (iv) (R) C) (IV) (ii) (P) D) (II) (i) (S)Key: DSol: Conceptual

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PART-III : MATHEMATICSSECTION 1(Maximum Marks:28)







37. If a chord, which is not a tangent, of the parabola 2 16y x has the equation 2x y p , and

midpoint (h,k), then which of the followin is (are) possible value(s) of p,h and k?A) p=-1,h=1,k=-3 B) p=5,h=4,k=-3 C) p=-2,h=2,k=-4 D) p=2,h=3,k=-4Key : D

Sol : 21 11 8 8 0 (1)S S x ky k h

but (1) is same as 2x+y-p=028 8

2 1

k k h

p

24, 4 8k P k h 4 16 8P h

4, 2 4k h p satisfied for D

38. Let a,b,x and y be real numbers such that a-b=1 and 0y . If the complex number

z x ty satisfies in1

az by

z

, then which of the following is (are) possible value (s) of x?

A) 21 1 y B) 21 1 y C) 21 1 y D) 21 1 y Key : B,D

Sol :( ) ( 1 )

2 1 1 ( 1) 1

az b a x iy b ax b aiy x iy

x iy x iy x iy

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2 2

( 1) ( )

1 ( 1)

az b ay x y ax by

Z x y

2 2 2 1a b x b x

2 2 2 0x y x

22 4 4

2

yx

21 1 y

39. Let X and Y be two events such that 1 1( ) , /

3 2P X P X Y and 2

/5

P X Y .Then

A) 2

5P X Y B) 4

5P Y C) 1

/2

P X Y D) 1

5P X Y

Key : B,C

Sol : , 2x y Any events

1 1 2( ) , ,

3 2 5

x yp x p p

y x

1 2 2 1 2( ), ( ) ( )

2 5 5 3 15p x y p y p x y p x

( ) ( )p x y p x p y p x y

1 4 2 5 4 2

3 15 15 15

2 1 4( ) ( )

15 2 15p x p y

7

15

also ( ) ( )

( ) ( )

p x yx p y p x yp

y p y p y

4 22 115 15

4 4 215

40. If 2x-y+1=0 is a tangent to the hyperbola2 2

21

16

x y

a , then which of the following CANNOT be

sides of a right angled triangle?A) 2a,4,1 B) a,4,1 C) a,4,2 D) 2a,8,1Key : B,C,D

Sol : 2 2 2 2 2 171 4 16

2c a m b a a

options (B),(C),(D) are not forming rt angled les41. Let [x] be the greatest integer less than or equals to x. Then at which of the following point(s) the

function f(x)= cos ( [ ])x x x is discontinuous?

A) x=2 B) x=0 C) x=1 D) x=-1

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Key : A,C,DSol : option (a)

(2) 2cos 4 2f lim

2 ( ) 2.cos(3 ) 2x f x x=2 is discent point option (b)f(0)=0lim lim

2 0( ) 0, ( ) 0x xf x f x x=0 cant pointoption (c)

(1) 1cos 2 1f lim

1 ( ) (1) cos( ) 1x f x x=+1 is discent pointoption (d)

( 1) ( 1)(1) 1f lim

1 ( ) ( 1) cos(3 ) 1x f x x=-1 is discent point

42. Which of the following is (are)NOT the square of a 3 3 matrix with real entries?

A)

1 0 0

0 1 0

0 0 1

B)

1 0 0

0 1 0

0 0 1

C)

1 0 0

0 1 0

0 0 1

D)

1 0 0

0 1 0

0 0 1

Key : A,DSol : Options (A),(B),(D) gives A with imaginary entries

43. Let : (0,1)f be a continuous function . Then which of the following function(s) has (have)the value zero at some point in the interval (0,1)?

A) 2

0( ) ( )sinf x f t tdt

B)0

( )sinxxe f t tdt C) 2

0( ) cos

xx f t tdt

D) 9 ( )x f x

Key : C,D

Sol :2

0

( ) ( ).cos

x

g x x f f t dt

0Let x

2

0

( ) 0 ( ).cosg x f f tdt

2

0

( ).cos ( )f f tdt ve

Let 9( ) ( )h x x f x

Let x=0 ( ) ( )h x f x Ve

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( ) (0,1)f x Let x=1

) 1 ( ) ( )hX f x Ve

( ) (0,1)f x

44. For a real number , if the system

2

2

1 1

1 1

1 1

x

y

z

of linear equations, has infinitely

many solutions, then 21 0 Key : 1

Sol :

2 2

2 3

2 2

1 1 1 1

1 1 0 1 1

1 1 0 0 1 1A B

infinite soltions 21 0,1 0 1

now 21 1

45. For how many values of p, the cirlce 2 2 2 4 0x y x y p and the coordinate have exactly

three common points?Key : 2Sol : case ( 1)

rad= 2

5 2p

1p

but r<d 5 1 4p p now p<-4 and p=-1 is not possiblecase (2)

5 1r d p

4p also we get p<-1 as in cae ..(1)case (3)

also 2 20 2 4 0p x x y y

cirlce through origin ( 1) ( 4) 0x x y y ( 0, 1), ( 0, 4)x y 2 possibilities

46. The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24 thenwhat is the length of its smallest side?Key : 6Sol : sides (a-d),(a),(a+d)

pyth thum 2 2 2 2( ) 2 2a d a d a ad ad a

24 4ad a a d thus sides ( ), , ( ) 3 , 4 ,5a d a a d d d d

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area of21

(3 )(4 ) 6 (24)( )2

le d d d given

2 4 2d d sides 3 , 4 ,5 6,8,10d d d

length of smallest side=647. Words of length 10 are fomred using the letters A,B,C,D,E,F,G,H,I,J. Let x be the number of

such words where no letter is repeated; and let y be the number of such words where exactly one

letter is repeated twice and no other letter is repeated. Then ,9

y

x

Key : 1

Sol :1 8

10!10! 10 9

2!C Cx and y 59

y

x

48. Let :f be a differentiable function such that (0) 0, 32

f f

and ' (0) 1f .If

2'( ) ( ) cos cot cos ( )

x

g x f t ec t t ec t f t dt

for 0,2

x

, then 0lim ( )x g x

Key : 5

Sol :'(0) 0; 3; (0) 1

2f f f

'2( ) [ ( ) cos cot cos . ( )]x

f x f t ect t ect t t dt

2 2

2cos . ( ) cos cot . ( ) cos cot . ( )x

x x

ect f t ect t f t dt ect t f t dt

cos . cos . ( )2 2

ec f ecx f x

( )3

sin

f x

x

now 0 0

( )( ) 3

sinlt ltx x

f xg x

x

0

( )3

sinltx

f x

x

'

0

( )3

cosltx

f x

x

13 3 1 2

1

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Columns 1,2 and 3 contain conics, equations of tangents to the conics and points of contactrespectivelyColum n 1 Colum n 2 Colum n 3

1) 2 2 2x y a i) 2my m x a (P)2

2,

a a

m m

(II) 2 2 2 2x a y a (ii) 2 1y mx a m (Q)22

,11

ma a

mm

(III) 2 4y ax (iii) 2 2 1y mx a m (R)2

2 2 2 2

1,

1 1

a m

a m a m

(IV) 2 2 2 2x a y a (iv) 2 2 1y mx a m (S)

2

2 2 2 2

1,

1 1

a m

a m a m

49. If a tangent to a suitable conic ( Column 1) is found to be 8y x and its point of contact is (8,16), then which of the following options is the only CORRECT combination?A)(III) (ii) (Q) B) (III) (i) (P) C) (II) (iv) (R) D) (I) (ii) (Q)Key : B

Sol : 2 4y ax Equation of tangenta

y mxm

given 8y x m = 1, c= 8

buta

cm

8 81

a a

m

8a

2

29, 8,16

a

m m

50. For 2,a if a tangent is derawn to a suitable conic (Column 1) at the point of contact (-1,1),

then which of the following options is the only CORRECCT combination for obtaining its equation?A) (I) (ii) (Q) B) (III) (i) (Q) C) (II) (ii) (Q) D) (III) (i) (P)

Key : A

Sol :2 2

, .a m b

c c

2 2 2c a x b

22 1c x

2

2 2

1,

2 1 2 1

a m

m x

51. The tangent to a suitable conic ( Column 1) at1

3,2

is found to be 3 2 4,x y then which of

the following options is the only CORRECT combination?A) (IV) (iii) (S) B) (II) (iv) (R) C) (II) (iii) (R) D) (IV)(iv) (S)

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Key : B

Sol :2 2

21 (1)

1

x y

a

2 2 1 (2)y mx a m

32

2y x

3, 2

2m c

2 2 2 2c a m b 2 4 2a a

Let log log , 0,e ef x x x x x x

Column 1 contains information about zeros of ' ".f x f x and f x

Column 2 contains information about the limiting behavior of ' ",f x f x and f x at infinity

Column 3 contains information about increasing decreasing nature of 'f x and f x

Column 1 Column 2 Column 3

(I) 20 1,f x for some x e (i) lim 0x f x (P) f is increasing in (0,1)

(II) ' 0 1,f x for some x e (ii) lim x f x (Q) f is decreasing in ( e,e2)

(III) ' 0 0,1f x for some x 'lim x f x (R) f’ is increasing in ( 0.1)

(IV) " 0 1,f x for some x e "lim 0x f x (S) f’ is decreasing in (e,e2)

52. Which of the following options is the only CORRECT combination?A) (IV) (iv) (S) B) (II) (ii) (Q) C) (I) (i) (P) D) (III) (iii) (R)Key : B) (II) (ii) (Q)

Sol : logn nf x x log e x e

' 1 11 logx

ef x xx x

1l o g x

ex

"2

1 1f x

x x

(I) 1 1 0f

2 22 0 ( )f e e T

( )II ' 1 1 0f

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' 11 0f e T

e

( )III ' 0f

' 1 1f F

( )IV " 1 2f

"2

1 1f e

e e (F)

(i) log log 0 ( )x x

xLt x e x e F

(ii) ( )xLt f x T

(iii) '

xLt f x T

(iv) " 0xLt f x T

(P) ' 0 ( )f x T

(Q) ' 210 log 0 , ( )

4f x x e e T

(R) "2

10 0 0,1 ( )

xf x F

x

(S)

"

22

0

10 , ( )

f x

xe e T

x

53. Which of the following options is the only INCORRECT combination ?A) (II) (iii) (P) B) (II) (iv) (Q) C) (I) (iii) (P) D) (III) (i) (R)Key : D) (III) (i) (R)


' 1 11 logx

ef x xx x

1l o g x

ex

"2

1 1f x

x x

(I) 1 1 0f

2 22 0 ( )f e e T

( )II ' 1 1 0f

' 11 0f e T

e

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( )III ' 0f

' 1 1f F

( )IV " 1 2f

"2

1 1f e

e e (F)


xLt x e x e F

(ii) ( )xLt f x T

(iii) '

xLt f x T

(iv) " 0xLt f x T

(P) ' 0 ( )f x T

(Q) ' 210 log 0 , ( )

4f x x e e T

(R) "2

10 0 0,1 ( )

xf x F

x

(S)

"

22

0

10 , ( )

f x

xe e T

x

54. Which of the following options is the only CORRECT combination ?A) (II) (iii) (S) B) (I) (ii) (R) C) (III) (iv) (P) D) (IV) (i)(S)Key : A


' 1 11 logx

ef x xx x

1l o g x

ex

"2

1 1f x

x x

(I) 1 1 0f

2 22 0 ( )f e e T

( )II ' 1 1 0f

' 11 0f e T

e

( )III ' 0f

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' 1 1f F

( )IV " 1 2f

"2

1 1f e

e e (F)


xLt x e x e F

(ii) ( )xLt f x T

(iii) '

xLt f x T

(iv) " 0xLt f x T

(P) ' 0 ( )f x T

(Q) ' 210 log 0 , ( )

4f x x e e T

(R) "2

10 0 0,1 ( )

xf x F

x

(S)

"

22

0

10 , ( )

f x

xe e T

x

*****

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JEE ADVANCED - 2017 Paper - I SOLUTIONSsmartprep.in/wp-content/uploads/2017/05/JEE...21 May 2017 JEE...

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