physics - m.media-amazon.com...9 8 1 rt (b) 3 2 1 rt (c) 15 8 1 rt (d) 9 2 1 rt ... an electron and...

Paper I

SECTION I

Directions (Q. Nos. 1-7) This section contains 7 multiple choice questions. Each question has

four choices (a), (b), (c) and (d) out of which only one is correct.

Consider an electric field E = E0 $x where E−

0 is a constant.

The flux through the shaded area ( as shown in thefigure) due to this field is(a) 2 0

2E a (b) 2 02E a

(c) E a02 (d)

E a02

2

A ball of mass ( )m 0.5 kg is attached to the end of a string havinglength ( )L 0.5 m. The ball is rotated on a horizontal circular pathabout vertical axis. The maximum tension that the string canbear is 324 N. The maximum possible value of angular velocityof ball (in rad/s) is(a) 9 (b) 18 (c) 27 (d) 36

A 2µF capacitor is charged as shown in the figure. Thepercentage of its stored energy dissipated after the switch S isturned to position 2, is(a) 0% (b) 20%(c) 75% (d) 80%

5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking theinitial temperature to be T1, the work done in the process is

(a)98 1RT (b)

32 1RT (c)

158 1RT (d)

92 1RT

A police car with a siren of frequency 8 kHz is moving with uniform velocity36 km/h towards a tall building which reflects the sound waves. The speed ofsound in air is 320 m/s. The frequency of the siren heard by the car driver is(a) 8.50 kHz (b) 8.25 kHz(c) 7.75 kHz (d) 7.50 kHz

The wavelength of the first spectral line in the Balmer series of hydrogen atom is6561 Å. The wavelength of the second spectral line in the Balmer series of singlyionized helium atom is(a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å

L

m

Z

Y

X(0,0,0) (0, ,0)a

( ,0, )a a ( , , )a a a

IIT JEESOLVED PAPER 2011

Physics

2 Fµ 8 Fµ–

1 2

A meter bridge is set up as shown in figure, todetermine an unknown resistance X using a standard10 Ω resistor. The galvanometer shows null pointwhen tapping key is at 52 cm mark. Theend-corrections are 1 cm and 2 cm respectively for theends A and B. The determined value of X is(a) 10.2 Ω (b) 10.6 Ω (c) 10.8 Ω (d) 11.1 Ω

SECTION II


four choices (a), (b), (c) and (d) out of which one or more may be correct.

An electron and a proton are moving on straight parallel paths with samevelocity. They enter a semi-infinite region of uniform magnetic fieldperpendicular to the velocity. Which of the following statement(s) is/are true?(a) They will never come out of the magnetic field region(b) They will come out travelling along parallel paths(c) They will come out at the same time(d) They will come out at different times

A composite block is made of slabs A B C D, , , and E

of different thermal conductivities (given in terms of aconstant, K) and sizes (given in terms of length, L) asshown in the figure. All slabs are of same width. HeatQ flows only from left to right through the blocks.Then, in steady state(a) heat flow through A and E slabs are same(b) heat flow through slab E is maximum(c) temperature difference across slab E is smallest(d) heat flow through C = heat flow through B + heat flow through D

A spherical metal shell A of radius R A and a solid metal sphere B of

radius R RB A< ( ) are kept far apart and each is given charge +Q. Now, they areconnected by a thin metal wire. Then(a) E A

inside = 0 (b) Q QA B>

(c)σσ

A

B

B

A

R

R= (d) E EA B

on surface on surface<

A metal rod of length L and mass m is pivoted at one end. A thindisc of mass M and radius R (< L) is attached at its centre to thefree end of the rod. Consider two ways the disc is attached. CaseA—the disc is not free to rotate about its centre and case B—thedisc is free to rotate about its centre. The rod-disc system performsSHM in vertical plane after being released from the same displaced position.Which of the following statement(s) is/are true?

2 IIT JEE (Physics) • Solved Paper 2011

1L

5L 6L

3L

4L

Heat0

A

2K 6K

EB

D

3K

5K

C 4K

x 10Ω

A B

(a) Restoring torque in case A = Restoring torque in case B

(b) Restoring torque in case A < Restoring torque in case B

(c) Angular frequency for case A > Angular frequency for case B

(d) Angular frequency for case A < Angular frequency for case B

SECTION III

Directions (Q. Nos. 12-16) This section contains 2 paragraphs. Based upon one of the

paragraph 2 multiple choice questions and based on the other paragraph 3 multiple choice

questions have to be answered. Each of these questions has four choices (a), (b), (c) and (d) out of

which only one is correct.

Paragraph (Q. Nos. 12-13) A dense collection of equal number of electrons and positive ions is

called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons

can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m.

When the electrons are subjected to an electric field, they are displaced relatively away from the

heavy positive ions.If the electric field becomes zero, the electrons being to oscillate about the

positive ions with a natural angular frequency ωp, which is called the plasma frequency. To

sustain the oscillations, a time varying electric field needs to be applied that has an angular

frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ωapproaches

ωp, all the free electrons are set to resonance together and all the energy is reflected. This is the

explanation of high reflectivity of metals.

Taking the electronic charge as e and the permittivity as ε0, use dimensionalanalysis to determine the correct expression for ωp .

(a)Ne

mε0

(b)m

Ne

ε0 (c)Ne

m

2

0ε(d)

m

Ne

ε02

Estimate the wavelength at which plasma reflection will occur for a metal havingthe density of electrons N ≈ × −4 1027 1

m . Take ε01110= − and m = −10 30,

where these quantities are in proper SI units.(a) 800 nm (b) 600 nm (c) 300 nm (d) 200 nm

Paragraph (Q. Nos. 14-16) Phase space diagrams are useful tools in

analyzing all kinds of dynamical problems. They are especially useful in

studying the changes in motion as initial position and momentum are

changed. Here we consider some simple dynamical systems in

one-dimension. For such systems, phase space is a plane in which position is

plotted along horizontal axis and momentum is plotted along vertical axis.

The phase space diagram is x t( ) vs p t( ) curve in this plane. The arrow on the

curve indicates the time flow. For example, the phase space diagram for a particle moving with

constant velocity is a straight line as shown in the figure. We use the sign convention in which

position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

The phase space diagram for a ball thrown vertically up from ground is

IIT JEE (Physics) • Solved Paper 2011 3

Position

Mo

me

ntu

m

Position

Momentum MomentumMomentumMomentum

(a) (c)(b) (d)Position

Position

Position

The phase space diagram for simple harmonic motion isa circle centered at the origin. In the figure, the twocircles represent the same oscillator but for differentinitial conditions, and E1 and E2 are the totalmechanical energies respectively. Then,(a) E E1 22= (b) E E1 22=(c) E E1 22= (d) E E1 216=

Consider the spring-mass system, with the mass submerged in water, as shownin the figure. The phase space diagram for one cycle of this system is

SECTION IV

Directions (Q. Nos. 17-23) This section contains 7 questions. The answer to each question is a

single digit integer, ranging from 0 to 9.

Four point charges, each of + q, are rigidly fixed at the four corners of a squareplanar soap film of side a. The surface tension of the soap film is γ. The system of

charges and planar film are in equilibrium, and a kq

N

=

2 1

γ

/

, where k is a

constant. Then, N is

A block is moving on an inclined plane making an angle 45° with the horizontaland the coefficient of friction is µ. The force required to just push it up theinclined plane is 3 times the force required to just prevent it from sliding down. Ifwe define N = 10 µ, then N is

A long circular tube of length 10 m and radius 0.3 m carries a currentI along its curved surface as shown. A wire loop of resistance 0.005 Ωand of radius 0.1 m is placed inside the tube with its axis coincidingwith the axis of the tube. The current varies as I I= 0 cos 300 t, whereI 0 is constant. If the magnetic moment of the loop is N I tµ 0 0 300sin ,then N is


Momentum

Position

Momentum

Position

Momentum

Position

Momentum

Position

(a) (c)(b) (d)

⇒I

E2

2a

a

Momentum

Position

E1

Steel wire of length L at 40° C is suspended from the ceiling and then a mass m ishung from its free end. The wire is cooled down from 40°C to 30°C to regain itsoriginal length L. The coefficient of linear thermal expansion of the steel is10 5− °/ C, Young’s modulus of steel is 1011 N/m 2 and radius of the wire is 1 mm.Assume that L >> diameter of the wire. Then, the value of m in kg is nearly

Four solid spheres each of diameter 5 cm and mass 0.5 kg ar placed with theircentres at the corners of a square of side 4 cm. The moment of inertia of thesystem about the diagonal of the square is N × −10 4 kg-m 2, then N is

A boy is pushing a ring of mass 2 kg and radius 0.5 m with astick as shown in the figure. The stick applies a force of 2 Non the ring and rolls it without slipping with an acceleration of0.3 m s/ 2. The coefficient of friction between the ground andthe ring is large enough that rolling always occurs and the

coefficient of friction between the stick and the ring isP

10. The

value of P is

The activity of a freshly prepared radioactive sample is 1010 disintegrations persecond, whose mean life is 10 9− s. The mass of an atom of this radioisotope is10 25− kg. The mass (in mg) of the radioactive sample is

Paper II



A light ray travelling in glass medium is incident on glass-air interface at an angleof incidence θ. The reflected ( )R and transmitted ( )T intensities, both as functionof θ, are plotted. The correct sketch is

A wooden block performs SHM on a frictionless surface withfrequency ν0. The block carries a charge + Q on its surface.If now a uniform electric field E is switched on as shown,then the SHM of the block will be(a) of the dame frequency and with shifted mean position(b) of the same frequency and with the same mean position(c) of changed frequency and with shifted mean position


Ground

Stick

(b)

Inte

nsity

100%T

R

0 θ 90°

Inte

nsity

(a)

100%T

R

0 θ 90°

(d)

Inte

nsity

100%T

R

0 θ 90°

Inte

nsity

100%T

R

0 θ 90°

(c)

+Q

E

(d) of changed frequency and with the same mean position

The density of a solid ball is to be determined in an experiment. The diameter ofthe ball is measured with a screw gauge, whose pitch is 0.5 mm and there are50 divisions on the circular scale. The reading on the main scale is 2.5 mm andthat on the circular scale is 20 divisions. If the measured mass of the ball has arelative error of 2%, the relative percentage error in the density is(a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2%

A ball of mass 0.2 kg rests on a vertical post of height5 m. A bullet of mass 0.01 kg, travelling with a velocityv m/s in a horizontal direction, hits the centre of the ball.After the collision, the ball and bullet travelindependently. The ball hits the ground at a distance of20 m and the bullet at a distance of 100 m from the footof the post. The initial velocity v of the bullet is(a) 250 m/s (b) 250 2 m/s (c) 400 m/s (d) 500 m/s

Which of the field patterns given in the figure is valid for electric field as well asfor magnetic field?

A point mass is subjected to two simultaneous sinusoidal displacements in

x-direction, x t A t1 ( ) sin= ω and x t A t223

( ) sin= +

ω π. Adding a third

sinusoidal displacement x t B t3 ( ) sin ( )= +ω φ bring the mass to a completerest. The values of B and φare

(a) 234

A,π

(b) A,43π

(c) 356

A,π

(d) A,π3

A long insulated copper wire is closely wound as a spiral of Nturns. The spiral has inner radius a and outer radius b. Thespiral lies in the XY-plane and a steady current I flows throughthe wire. The Z-component of the magnetic field at the centreof the spiral is

(a)µ 0

2NI

b a

b

a( )ln

−

(b)µ 0

2NI

b a

b a

b a( )ln

−+−

(c)µ 0

2NI

b

b

aln

(d)µ 0

2NI

b

b a

b aln

+−

A satellite is moving with a constant speed v in a circular orbit about the earth.An object of mass m is ejected from the satellite such that it just escapes from thegravitational pull of the earth. At the time of its ejection, the kinetic energy of theobject is


(a) (b) (c) (d)

v m/s

0 20 m 100 m

a

bI

Y

X

(a)12

2mv (b) mv2 (c)32

2mv (d) 2 2mv

SECTION II



Two solid spheres A and B of equal volumes but of differentdensities d A and dB are connected by a string. They are fullyimmersed in a fluid of density dF . They get arranged into anequilibrium state as shown in the figure with a tension in thestring. The arrangement is possible only if(a) d dA F< (b) d dB F>(c) d dA F> (d) d d dA B F+ = 2

Which of the following statement(s) is/are correct?(a) If the electric field due to a point charge varies as r−2 5. instead of r−2, then the

Gauss’s law will still be valid(b) The Gauss’s law can be used to calculate the field distribution around an electric

dipole(c) If the electric field between two point charges is zero somewhere, then the sign of

the two charges is the same(d) The work done by the external force in moving a unit positive charge from point

A at potential VA to point B at potential VB is ( )V VB A−

A series R-C circuit is connected to AC voltage source. Consider two cases; ( )A

when C is without a dielectric medium and (B) when C is filled with dielectric ofconstant 4. The current I R through the resistor and voltage VC across thecapacitor are compared in the two cases. Which of the following is/are true?(a) I IR

ARB> (b) I IR

ARB< (c) V VC

ACB> (d) V VC

ACB<

A thin ring of mass 2 kg and radius 0.5 m is rollingwithout slipping on a horizontal plane with velocity 1m/s. A small ball of mass 0.1 kg, moving with velocity20 m/s in the opposite directions, hits the ring at aheight of 0.75 m and goes vertically up with velocity 10m/s. Immediately after the collision(a) the ring has pure rotation about its stationary CM(b) the ring comes to a complete stop(c) friction between the ring and the ground is to the left(d) there is no friction between the ring and the ground

SECTION III

Directions (Q. Nos. 13-18) This sections contain 6 questions. The

answer to each of the question is a single-digit integer, ranging from 0to 9.


A

B

0.75 m

20 m/s

10 m/s

1 m/s

A B6 V

3 V 2 Ω

1 Ω

Two batteries of different emfs and different internal resistances are connectedas shown. The voltage across AB in volt is

A series R-C combination is connected to an AC voltage of angular frequencyω = 500 rad/s. If the impedance of the R-C circuit is R 125. , the time constant(in millisecond) of the circuit isA train is moving along a straight line with a constant acceleration a. A boystanding in the train throws a ball forward with a speed of 10 m/s, at an angle of60° to the horizontal. The boy has to move forward by 1.15 m inside the train tocatch the ball back at the initial height. The acceleration of the train in m/s 2, is

Water (with refractive index = 43

) in a tank is 18 cm deep.

Oil of refractive index74

lies on water making a convex

surface of radius of curvature R = 6 cm as shown.Consider oil to act as a thin lens. An object S is placed24 cm above water surface. The location of its image is at x cm above thebottom of the tank. Then, x is

A block of mass 0.18 kg is attached to a spring of forceconstant 2 N/m. The coefficient of friction between theblock and the floor is 0.1. Initially, the block is at rest andthe spring is unstretched. An impulse is given to the block as shown in the figure.The block slides a distance of 0.06 m and comes to rest for the first time. The

initial velocity of the block in m/s is vN=10

. Then, N is

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from aninsulating thread in free-space. It is under continuous illumination of 200 nmwavelength light. As photoelectrons are emitted, the sphere gets charged andacquires a potential. The maximum number of photoelectrons emitted from thesphere is A Z× 10 (where 1 10< <A ). The value of Z is

SECTION IV

Directions (Q. Nos. 19-20) This section contains 2 questions. Each question has four

statements (A, B, C and D) given in Column I and five statements (p,q, r, s and t) in Column II.Any given statement in Column I can have correct matching with one or more statement(s) given

in Column II.

One mole of a monatomic ideal gas is takenthrough a cycle ABCDA as shown in the p-Vdiagram. Column II gives the characteristicsinvolved in the cycle. Match them with each of theprocesses given in Column I.


R = 6 cm

S

µ = 7/4µ = 1.0

µ = 4/3

⇒

B A3p

1pC D

p

0 1V 3V 9VV

Column I Column II

(A) Process A B→ (p) Internal energy decreases(B) Process B C→ (q) Internal energy increases(C) Process C D→ (r) Heat is lost(D) Process D A→ (s) Heat is gained

(t) Work is done on the gas

Column I showns four systems, each of the same length L, for producingstanding waves. The lowest possible natural frequency of a system is called itsfundamental frequency, whose wavelength is denoted as λ f . Match each systemwith statements given in Column II describing the nature and wave length of thestanding waves.

Column I Column II

(A) Pipe closed at one end (p) Longitudinal waves

(B) Pipe open at both ends (q) Transverse waves

(C) Stretched wire clamped at both ends (r) λ f L=

(D) Stretched wire clamped at both ends and atmid-point

(s) λ f L= 2

(t) λ f L= 4

ANSWERS

Paper I1. (c) 2. (d) 3. (d) 4. (a) 5. (a) 6. (a)7. (b) 8. (b,d) 9. (a,c,d) 10. (a,b,c,d) 11. (a,d) 12. (c)

13. (b) 14. (d) 15. (c) 16. (b) 17. (3) 18. (5)19. (6) 20. (3) 21. (9) 22. (4) 23. (1)

Paper II1. (c) 2. (a) 3. (c) 4. (d) 5. (c) 6. (b)7. (a) 8. (b) 9. (a,b,d) 10. (c,d) 11. (b,c) 12. (a,c)

13. (5) 14. (4) 15. (5) 16. (2) 17. (4) 18. (7)19. (A) → p,r,t; (B) → p,r; (C) → q, s; (D) → r 20. (A) → p,t; (B) → p,s; (C) → q, s; (D) → q, r


O L

O L

O L

O L

L/2

Hints & Solutions

Paper I

1. Electric flux, E S⋅ , or φ θ= ES cos

Here, θ is the angle between E and S.In this question θ = °45 , because S is

perpendicular to the surface.E E= 0

⇒ S a a a= =( ) ( )2 2 2

∴ φ = ° =( ) ( ) cosE a E a02

022 45

∴ Correct option is (c).Analysis of Question

(i) Question is moderately tough.(ii) The given shaded area is a rectangle

not a square. One side of thisrectangle is a and other side is 2a .

(iii) Electric field is uniform, whosemagnitude is E0 and direction ispositive x. In uniform electric fieldwe can use, φ θ= ES cos

2.

r l= sin θT cos θ component will cancel mg .T sin θ component will providenecessary centripetal force to the balltowards centre C.∴ T mr m lsin ( sin )θ ω θ ω= =2 2

or T ml= ω2

∴ ω = T

ml

or ωmax = =×

T

ml

max324

0 5 0 5. .

= 36 rad/s∴ Correct option is (d).Analysis of Question

(i) Question is simple.

(ii) This is called the conical pendulum.(iii) The interesting fact in this problem is

that ω or T is independent of θ.ω ∝ T

If ω is increased, T will also increase.

3. q C V V qi i= = =2 (say)

This charge will remain constant afterswitch is shifted from position 1 toposition 2.

Uq

C

q qi

i

= =×

=12 2 2 4

2 2 2

Uq

Cf

f

= 12

2

=×

=q q2 2

2 10 20

∴ Energy dissipated = −U Ui f = q2

5

This energy dissipated =

q2

5is 80%

of the initial stored energy =

q2

4.

Analysis of Question

(i) This question is moderately tough.(ii) In a capacitor circuit, redistribution

of charge takes place underfollowing three conditions.(a) A switch is closed.(b) A closed switch is opened.(c) A switch is shifted from one

position to another position.In the redistribution of charge, energy isdissipated.

4. At STP, 22.4 L of any gas is 1 mole.

∴ 5.6 L = =5 6224

14

..

moles = n

In adiabatic process,TV γ − =1 constant

∴ T V T V2 21

1 11γ γ− −=

or T TV

V2 1

1

2

=

γ−1

θ

mgr

lθ

T

γ = =C

C

p

V

53

for monoatomic He gas.

∴ T T T2 1

5

31

15 607

4=

=

−..

Q Further in adiabatic process,Q = 0

∴ W U+ =∆ 0

or W U= − ∆= −nC TV∆

= −−

−n

RT T

γ 1 2 1( )

= −−

−14 5

31

4 1 1R

T T( )

= − 98 1RT

∴ Correct option is (a).Analysis of Question

(i) From calculation point of viewquestion is moderately tough.

(ii) Volume of gas is decreasing.Therefore, work done by the gasshould be negative.

5. 36 km/h−1 = ×365

18= 10 ms−1

Apparent frequency of sound heard bycar driver (observer) reflected from thebuilding will be

f fv v

v vs

′ = +−

0

= +−

8

320 10320 10

= 8 5. kHz∴ Correct option is (a).Analysis of Question


(ii) Driver will listen two sounds, directand reflected. Direct sound will be of8 kHz as driver has no relativemotion with the car. But reflectedsound is of increased frequencybecause driver and image of carboth are approaching towards eachother.

6.

For hydrogen or hydrogen type atoms

1 1 122 2λ

= −

RZ

n nf i

In the transition from n ni f→

∴ λ ∝

−

1

1 122 2Z

n nf i

∴ λλ

2

1

12

2 2

1

22

2 2

2

1 1

1 1=

−

−

Zn n

Zn n

f i

f i

λ

λ

2

1 12

2 2

1

22

2 2

2

1 1

1 1=

−

−

Zn n

Zn n

f i

f i

Substituting the values, we have

=−

−

( ) ( )

( )

6561 11

2

1

3

21

2

1

4

22 2

22 2

Å= 1215 Å

∴ Correct option is (a).Analysis of Question

(i) Question is simple.(ii) In modern physics, mostly questions

are asked on the emission of photonby the transition of electron fromsome higher energy state to somelower energy state.


car

Observer

10 ms–1

10 ms–1

car Source

Building For reflectedsound

n = 4 n = 4

n = 3 n = 3

n = 2 n = 2

n = 1 n = 2

First line ofBalmar series

Second line ofBalmar series

(iii) For hydrogen and hydrogen likeatoms which have only singleelectron, we can use the formula

1 1 122 2λ

= −

RZ

n nf i

(iv) Further, the student should alsoremember different series, as shownin figure below.

I → Lymen seriesII → Balmer series

III → Paschen seriesIV → Brackett seriesV → Pfund series

7. Using the concept of balancedWheatstone bridge, we have

P

Q

R

S=

∴ X

( ) ( )52 110

48 2+=

+

∴ X = × =10 5350

10 6. Ω

∴ Correct option is (b).Analysis of Question

Question is moderately tough, becausenormally end corrections are not taughtin the coaching/schools in this type ofproblem.

8. rmv

Bq=

or r m∝∴ r re p< as m me p<

Further, Tm

Bq= 2π

or T m∝

∴ T Te p<

tT

ee=2

and tT

pp=2

or t te p<∴ Correct options are (b) and (d).

Note In JEE 2011, official answer key,correct options were given bc, bd,bcd, here student should realise thatboth the electron and proton enterthe magnetic field at the same timeor at different times.


(i) Question is simple.(ii) In the situation given above particle

will never complete the circle unlessmagnetic field is all around.

9. Thermal resistance

Rl

KA=

∴ RL

K Lw KwA = =

( ) ( )2 41

8

(Here, w = width)

RL

K Lw KwB = =4

34

3( )

RL

K Lw KwC = =4

4 21

2( ) ( )

RL

K Lw KwD = =4

54

5( ) ( )

RL

K Lw KwE = =

( )( )61

6

R R R R RA B C D E: : : := 15 160 60 96 12: : : :

So, let us write, RA = =15 160R R RB,etc and draw a simple electrical circuitas shown in figure.


I

II

III

IVV

r

s

× × × ×

× × × ×

H = Heat current = Rate of heat flow.H H HA E= = [let]

∴ Option (a) is correct.In parallel, current distributes in inverseratio of resistance.

∴ H H HR R R

B C DB C D

: : : := 1 1 1

= 1160

160

196

: :

= 9 24 15: :

∴ H H HB =+ +

=9

9 24 153

16

H H HC =+ +

=24

9 24 1512

and H H HD =+ +

=15

9 24 155

16

H H HC B D= +∴ Option (d) is correct.Temperature difference (let us call it T)= (Heat current) × (Thermal resistance)

T H R H R HRA A A= = =( ) ( )15 15

T H R H R HRB B B= =

=3

16160 30( )

T H R H R HRC C C= =

=1

260 30( )

T H R H R HRD D D= =

=5

1696 30( )

T H R H R HRE E E= = =( ) ( )12 12Here, TE is minimum. Therefore, option(c) is also correct.∴ Correct options are (a), (c) and (d).Analysis of Question

(i) From calculation point of view,question is difficult otherwisequestion is simple.

(ii) In heat transfer, questions aremainly asked from conduction andradiation topic.

10. Inside a conducting shell, electric field isalways zero. Therefore, option (a) iscorrect. When the two are connected,their potentials become the same.∴ V VA B=

orQ

R

Q

R

A

A

B

B

=

Q VQ

R=

14 0πε

Since, R RA B>∴ Q QA B>∴ Option (b) is correct.Potential is also equal to,

VR= σ

ε0

V VA B=∴ σ σA A B BR R=

orσσ

A

B

B

A

R

R=

∴ Option (c) is correct.Electric field on surface,

EE

E= ∝σ σ0

or

or σ σA B<Since, σ σA B<∴ E EA B<∴ Option (d) is also correct.∴ Correct options are (a), (b), (c)and (d).Analysis of Question


(ii) EQ

R= =1

4 02

0πεσε

(On surface)

As, σ = surface charge density

= Q

R4 2π

(iii) VQ

R

R= =14 0 0πε

σε

11. τ τ θ θA B mgL

MgL= = +2

sin sin

= Restoring torque about point O.


15R

A

12R

160R

HB

96R

HC

B

H EH

HD D

60 R

In case A, moment of inertia will bemore. Hence, angular acceleration( / )α τ= I will be less. Therefore, angularfrequency will be less.

∴ Correct options are (a) and (d).Analysis of Question

Question is difficult from my point ofview. Because this type of SHM is rarelytaught in the class and questions of thistype are not given in standard books.

12. N = Number of electrons per unit

volume∴ [N] = =−[ ],[ ] [ ]L

3 e q

= =[ ] [ ]It AT

[ ] [ ]ε01 3= − −

M L T A4 2

Substituting the dimensions, we can seethat,

Ne

m

2

0

1

ε

= −[ ]T

Angular frequency has also thedimension [ ]T−1 .∴ Correct option is (c).Analysis of Question

(i) From calculation point of view,question is moderately difficult.Otherwise it is simple.

(ii) Students may commit a mistake inthe dimensions of N. It is notdimensionless.

13. ω = 2π πλ

fc= 2

∴ λ πω

π

ε= =2 2

20

c c

Ne m/

Substituting the values, we getλ = 600 nm

∴ Correct option is (b).


(i) Paragraph does not make thesolution very clear.

(ii) Solution is simple but based on errorand trial method.

14. Momentum is first positive butdecreasing. Displacement (or sayposition). is initially zero.It will first increase. At highest point,momentum is zero and displacement ismaximum. After that momentum isdownwards (negative) and increasingbut displacement is decreasing. Only (d)option satisfies these conditions.


(i) Question is simple.(ii) Since all the graphs given in

question are parabolic type, weneed not to check the mathematicalpart of the question.

15. E m A= 12

2 2ω

or E A∝ 2

E

E

A

A

2

1

2

1

2

=

=

a

a2

2

or E E1 24=∴ Correct option is (c).


(i) Question is simple.(ii) Since, oscillator is same, ω ω1 2=

16. In all the given four figures, at meanposition the position coordinate is zero.At the same time mass is starting fromthe extreme position in all fours cases. Infigures (c) and (d), extreme position ismore than the initial extreme position.But due to viscosity opposite should bethe case. Hence, the answer should beeither option (a) or option (b).Correct answer is (b), because massstarts from positive extreme position(from uppermost position). Then, it willmove downwards or momentum shouldbe negative.


Mg

mg

θ

O


Question is simple but (a) and ( )b

options are confusing.

17.

F1 = Net electrostatic force on anyonecharge due to rest of three charges

= +

14

2120

2

2πεq

a

F2 = Surface tension force = γa

If we see the equilibrium of line BC,

then, 2 451 2F Fcos ° =or 2 1 2F F=

or1

42

1

20

2

2πεγq

aa+

=

∴ aq3

0

214

21

2= +

πε γ

or aq= +

14

21

20

1 3 2 1 3

πε γ

/ /

=

k

q2 1 3

γ

/

where, k = +

14

21

20

1 3

πε

/

Therefore, N = 3Answer is 3.


(i) Question is moderately difficult.(ii) In my opinion problems of surface

tension or viscosity are always notvery simple questions.

18.

F mg mg1 = +sin cosθ µ θF mg mg2 = −sin cosθ µ θ

Given that, F F1 23=or (sin cos )45 45° + °µ

= ° − °3 45 45(sin cos )µOn solving, we get µ = 0 5.∴ N = =10 5µ∴ Answer is 5.Analysis of Question

Question is simple. Only one has to takecare of direction and magnitude offriction.

19. Take the circular tube as a longsolenoid. The wires are closely wound.Magnetic field inside the solenoid is

B ni= µ0

Here, n = number of turns per unitlength∴ ni = current per unit lengthIn the given problem,

niI

L=

∴ BI

L= µ0

Flux passing through the circular coil is

φ µ π= =

BS

I

Lr0 2( )

Induced emf, ed

dt

r

LR

dI

dt= − = −

φ µ π02

.

Induced current,

ie

R

r

LR

dI

dt= = −

µ π02

Magnetic moment iA i r= π 2


F1

F1 F1

F1F1

A B

CD

45°

45°

θ

F1 f

Moving upwords

q

F2 f θ = 45°

Just remains stationary

or Mr

LR

dI

dt= −

⋅µ π0

2 4

…(i)

Given, I I t= 0 300cos

∴ dI

dtI t= − 300 3000 sin ( )

Substituting in Eq. (i), we get

Mr

LRI t=

300300

2 4

0 0π µ sin

∴ Nr

LR= 300 2 4π

Substituting the values, we get

N = 300 22 7 0110 0 005

2 4( / ) ( . )( )( . )

= 5 926.or N ~− 6


Question is difficult to understand insidethe examination hall. But 5 to 10%question in IIT JEE are always difficult.Students should not panic. Becausetopper of IIT JEE scores approximately80-90%.

20. ∆lFL

AY

mgL

r Y1 2= = =

πIncrease in length

∆ ∆l L2 = =α θ Decrease in lengthTo regain its original length,

∆ ∆l l1 2=

∴ mgL

r YL

πα θ2 = ∆

∴ mr Y

g=

2 α θ∆


m ~− 3 kg

∴ Answer is 3.


Question is very simple. In my opinionany student who have gone through thesyllabus once can solve this problemeasily.

21. rd= =2

52

cm

= × −52

10 2 m

m = 0 5. kga = 4 cm

= × −4 10 2 mI I I I IXX = + + +1 2 3 4

= +

+2

5 2

25

22

2mr ma

mr

+ +

+2

5 2

25

22

2mr ma

mr

Substituting the values, we getIXX = × −9 10 4 Kgm−2

∴ N = 9Answer is 9.


(i) Question is simple.(ii) Only theorem of parallel axes is to

be used properly.

22.

There is no slipping between ring andground. Hence, f2 is not maximum. Butthere is slipping between ring and stick.Therefore, f1 is maximum. Now, let uswrite the equations.

I mR= =2 22 0 5( ) ( . )

= 12

kgm−2

N F ma1 2− =or N F1 2 2 0 3 0 6− = =( ) ( . ) . N …(i)


a/ 2√

a

4 3

21 X

X

mg

N1

f1

f2

a

N2

a RR

I= =α τ

= −R f f R

I

( )2 1 = −R f f

I

22 1( )

∴ 0 30 5

1 2

22 1.

( . ) ( )( / )

= −f f

or f f2 1 0 6− = . N …(ii)

N f12

12 22 4+ = =( ) …(iii)

Further F N1 1= µ =

PN

10 1 …(iv)

Solving above four equation, we getP ~ .− 3 6

Therefore, the correct answer shouldbe 4.


(i) Question is moderately tough fromconcept point of view. Butcalculations are lengthy.

(ii) One has to think about the twocomponents of the force applied bythe stick.

(iii) Answer comes an integer when youconsider only N1.

23. Activity −

= =

×dN

dtN

tNλ 1

mean

∴ NdN

dtt= −

× mean

= Total number of atomsMass of one atom is 10 25− kg = m (say)∴ Total mass of radioactive substance

= (number of atoms)× (mass of one atom)

= −

dN

dtt m( )( )mean

Substituting the values, we getTotal mass of radioactive substance

= 1 mg∴ Answer is 1.Analysis of Question

Question is very simple. Again in myopinion one can solve it.

Paper II

1. After critical angle reflection will be100% and transmission is 0%. Options(b) and (c) satisfy this condition. Butoption (c) is the correct option. Becausein option (b) transmission is given 100%at θ = °0 , which is not true.


From examination point of viewquestion is moderately difficult. Becauseit takes little time to understand thegiven graphs clearly.

2. Frequency or time period of SHMdepends on variable forces. It does notdepend on constant external force.Constant external force can only changethe mean position. For example, in thegiven question mean position is atnatural length of spring in the absenceof electric field. Whereas in the presenceof electric field mean position will beobtained after a compression of x0.Where x0 is given by

kx QE0 =

or xQE

k0 =

∴ Correct answer is (a).Analysis of Question


(ii) I think almost all the seriousaspirants of JEE can solve thisproblem easily.


3. Least count of screw gauge = 0.550

= 0 01. mm = ∆r

Diameter, r = 2 5. mm + ×20500.5

= 270. mm∆r

r= 0.01

2.70

or∆r

r× =100

12.7

Now, density dm

V

m

r= =

43 2

3

π

Here, r is the diameter.

∴ ∆ ∆ ∆d

d

m

m

r

r× = +

×100 3 100

= × + ×

×∆ ∆m

m

r

r100 3 100

= + ×2 3127

%.

= 311. %Analysis of Question

(i) Question is moderately difficult.(ii) In practical part, questions in JEE

are asked from vernier callipers andscrew gauge.

4. Time taken by the bullet and ball tostrike the ground is

th

g= = × =2 2 5

101 s

Let v1 and v2 are the velocities of balland bullet after collision.Then applying

x vt=We have, 20 11= ×v

or v1 20= ms−1

100 12= ×v or v2 100= m/s−1

Now, from conservation of linearmomentum before and after collisionwe have,

0 01 0 2 20 0 01 100. ( . ) ( . )v = × + ×On solving, we get

v = 500 ms−1

∴ Correct answer is (d).


Question is moderately lengthy fromcalculation point of view, otherwise it issimple.

5. Correct answer is (c), because inducedelectric field lines (produced by changein magnetic field) and magnetic fieldlines form closed loops.Analysis of Question

Question is simple, provided you havean idea of induced electric field lines.

6.

Resultant amplitude of x1 and x2 is A at

angleπ3

from A1. To make resultant of

x x1 2, and x3 to be zero. A3 should be

equal to A at angle φ π= 43

as shown in

figure.∴ Correct answer is (b).

Alternate Solution

If we substitute,x x x1 2 3 0+ + =

or A t A tsin sinω ω π+ +

23

+ + =B tsin ( )ω φ 0Then, by applying simple mathematics,we can prove that

B A=

and φ π= 43


(i) Question is simple.(ii) Question can be solved by applying

mathematics also.(iii) Amplitudes of two or more sine or

cosine functions of same frequencyω can be added by vector method.


A A2 =A

A A1 =4 /3p

A A3 =

2 /3p

7. If we take a small strip of dr at distance r

from centre, then number of turns in thisstrip would be

dNN

b adr=

−Magnetic field due to this element atthe centre of the coil will be

dBdN I

r= µ0

2( ) =

−µ0

2NI

b a

dr

r( )

∴ B dBNI

b a

b

ar a

r b= =

−=

=

∫µ0

2( )ln

∴ Correct answer is (a).


(i) If we see this problemindependently, then I will rate thisquestion moderately difficult. Butthe idea of this question is takenfrom question number 3.245 of IEIrodov.

(ii) Interestingly the same question wasasked in IIT-JEE 2001 also.

8. In circular orbit of a satellite, potentialenergy

= − ×2 (kinetic energy)

= − × = −212

2 2mv mv

Just to escape from the gravitationalpull, its total mechanical energy shouldbe zero. Therefore, its kinetic energyshould be + mv2.∴ Correct answer is (b).Analysis of Question

(i) Question is moderately difficult.(ii) In circular orbit of satellite

UGMm

r= − and K

GMm

r=

2or U K= − 2Here, U is potential energy and K iskinetic energy.

9.

F = Upthrust = Vd gF

Equilibrium of A

Vd g T WF A= += +T Vd gA …(i)

Equilibrium of B,T Vd g Vd gF B+ = …(ii)

Adding Eqs. (i) and (ii), we get2d d df A B= +

∴ Option (d) is correct.From Eq. (i), we can see that

d dF A> [as T > 0]∴ Option (a) is correct.From Eq. (ii) we can see that,

d dB F>∴ Option (a) is correct.∴ Correct options are (a), (b) and (d).Analysis of Question

Question is moderately difficult butconceptwise it is good.

10. If charges are of opposite signs, then thetwo fields are along the same direction.So, they cannot be zero. Hence, thecharges should be of same sign.Therefore, option (c) is correct.Further, work done by external force =change in potential energy∴ W q VA B→ = ( )∆

= + −( ) ( )1 V VB A

or W V VA B B A→ = −Therefore, option (d) is also correct.∴ Correct options are (c) and (d).


(i) Question is simple.(ii) W U= ∆ (by external force)


A

B

F

WA

T

T F

WB

If charge is just moved, withoutchanging the kinetic energy. If nothing ismentioned in the question regarding thekinetic energy then we take W U= ∆ .

11. Z R X RC

C= + = +

2 2 221

ωIn case (b) capacitance C will be more.Therefore, impedance Z will be less.Hence, current will be more.∴ Option (b) is correct.

Further, V V VC R= −2 2

= −V IR2 2( )

In case (b), since current I is more.Therefore, VC will be less.∴ Option (c) is correct.∴ Correct options are (b) and (c).


(i) Question is moderately difficult.(ii) In my opinion problems of

alternating currents are not verydifficult.

(iii) Topic of AC is small. One can feelcomfortable in this topic by puttingless efforts.

12. The data is incomplete. Let us assumethat friction from ground on ring is notimpulsive during impact.From linear momentum conservation inhorizontal direction, we have

( ) ( . )− × + ×2 1 01 20

= × + ×( . ) ( )01 0 2 v ← →− +ve ve

Here, v is the velocity of CM of ring afterimpact.Solving the above equation, we have

v = 0Thus, CM becomes stationary.∴ Correct answer is (a).Linear impulse during impact

(i) In horizontal directionJ P1 01 20 2= = × =∆ . Ns

(ii) In vertical directionJ P2 01 10 1= = × =∆ . Ns

Writing the equation (about CM)Angular impulse

= Change in angular momentum

132

12

2 0 512

× ×

− × ×.

= × −

2 0 51

0 52( . )

.ω

Solving this equation ω comes out to bepositive or ωanti-clockwise. So just aftercollision rightwards slipping is takingplace.Hence, friction is leftwards.Therefore, option (c) is also correct.∴ Correct options are (a) and (c).

Note In JEE 2011 official answer key,correct option were given a, ac.


(i) Question is moderately difficult.(ii) In such type of problems impulse

due to friction during collision isignored.

13. VAB = Equivalent emf of two batteries in

parallel.

= ++

E r E r

r r

1 1 1 2

1 21 1/ // /

= ++

( / ) ( / )( / ) ( / )6 1 3 21 1 1 2

= 5 V

∴ Answer is 5.Analysis of Question

(i) In my opinion this is the simplestquestion of JEE 2011.

(ii) When no current is drawn from thisequivalent battery, then

V V EAB = =Otherwise, V E ir= ±

(iii) Similar type of question was askedin JEE 1981 and cancelled paper ofJEE 1997.

14. Z R XC= +2 2 = R 1 25.


w

1 N s

2 N s

+ve –ve

∴ R X RC2 2 21 25+ = .

or XR

C =2

or1

2ωC

R=

∴ Time constant = =CR2ω

= 2500

s = 4 ms

∴ Answer is 4.Analysis of Question

(i) Question is very simple.(ii) I think this is one of the simplest

formula based question of thispaper.

15. t Tu

g= = 2 sin θ

= × × ° =2 10 6010

3sin

s

Displacement of train in time t at= 12

2

Displacement of boy with respect totrain

= 115. m∴ Displacement of boy with respect to

ground = +

115

12

2. at

Displacement of ball with respect toground = °( cos )u t60To catch the ball back at initial height,

11512

602. ( cos )+ = °at u t

∴ 11512

3 1012

32. ( )+ = × ×a

Solving this equation, we geta = 5 ms−2

∴ Answer is 5.


(i) Question is moderately tough.(ii) Velocity of ball given in the question

is with respect to ground.

16.

Two refractions will take place, first fromspherical surface and the other from theplane surface.So, applying

µ µ µ µ2 1 2 1

v u R− = −

two times with proper sign convention.Ray of light is travelling downwards.Therefore, downward direction is takenas positive direction.

7 4 1 024

7 4 1 06

/ . / .v

−−

= −+

…(i)

4 318

7 4 4 3 7 4/( )

/ / /−

− = −∝x v

…(ii)

Solving these equations, we getx = 2 cm

∴ Answer is 2.


(i) Question is moderately difficult fromcalculation point of view, otherwiseit is simple.

(ii)µ µ µ µ2 1 2 1

v u R− = −

can be applied

forplane surface also with R = ∝

17. Decrease in mechanical energy= Work done against friction

∴ 12

12

2 2mv kx mgx− = µ

or vmgx kx

m= +2 2µ


v = 0 4. ms−1 =

410

ms−1

∴ Answer is 4.




24 cmm = 7/4

m =4—3 18 cm

xI

S

(ii) If µ s and µk two values of coefficientof friction are given, then we willtake µk.

18. Photoemission will stop when potentialon silver sphere becomes equal to thestopping potential.

∴ hcW eV

λ− = 0

Here, Vne

r0

0

14

=πε

∴ 12401200

4 7eV eV

− ( . )

= × × × × −

−9 10 1 6 10

10

9 19

2

n .

( . . ).

6 2 4 79 10 1 6 10

10

9 19

2− = × × × × −

−n

or n = ×× ×

−

−1 5 10

9 1 6 10

2

10

.

.= ×1 04 107.

∴ Answer is 7.


(i) Question is moderately difficult.(ii) In eV0 if we substitute the value of

e = × −1 6 10 19. C, then answer is inJ. If we leave it then, answer is in eV.

19. Internal energy ∝ ∝T pV

This is because

Unf

RT=2

= fpV

2Here, n = number of moles

f = degree of freedomIf the product pV increases, theninternal energy will increase and ifproduct decreases, the internal energywill decrease.Further, work is done on the gas, ifvolume of gas decreases. For heatexchange.

Q W U= + ∆

Work done is area under p-V graph. Ifvolume increases work done by gas ispositive and if volume decreases workdone by gas is negative. Further ∆U ispositive if product of pV is increasingand ∆U is negative, if product of pV isdecreasing. If heat is taken by the gas Q

is positive and if heat is lost by the gas Q

is negative.Keeping the above points in mind theanswer to this question is as under.

(A) → (p, r, t)(B) → (p, r)(C) → (q, s)(D) → (r, t)


(i) Calculation wise, question is slightlylengthy. Otherwise question istheory based and simple.

(ii) In process DA,p V p VA A D D=

∴ T TA D=or ∆U = 0Further, volume of gas isdecreasing. Therefore, work is doneon the gas or work done by gas isnegative. Therefore, Q is negative orheat is lost.

(iii) This question covers almost all theconcepts of first law ofthermodynamics.

20. In organ pipes, longitudinal waves areformed. In string, transverse waves areformed. Open end of pipe isdisplacement antinode and closed endis displacement node. In case of stringfixed end of a string is node.Further, least distance between a node

and an antinode isλ4

and between two

nodes isλ2

. Keeping these points in

mind answer to this question is asunder;(A) → (p, t)(B) → (p, s)(C) → (q, s)


Paper I

SECTION I



Let a i j k= + +$ $ $ , b i j k= − +$ $ $ and c i j k= − −$ $ $ be three vectors. A vector v in

the plane of a and b whose projection of c is1

3, is given by

(a) $ $ $i j k− +3 3 (b) − − −3 3$ $ $i j k

(c) 3 3$ $ $i k− +j (d) $ $ $i k k+ −3 3

Let the straight line x b= divide the area enclosed by y x= −( )1 2, y = 0 and

x = 0 into two parts R x b1 0( )≤ ≤ and R b x2 1( )≤ ≤ such that R R1 21

4− = .

Then, b equals to

(a)3

4(b)

1

2(c)

1

3(d)

1

4

A straight line L through the point ( , )3 2− is inclined at an angle 60° to the line

3 1x y+ = . If L also intersects the X-axis, then the equation of L is

(a) y x+ + − =3 2 3 3 0 (b) y x− + + =3 2 3 3 0

(c) 3 3 2 3 0y x− + + = (d) 3 3 2 3 0y x+ − + =

The value ofx x

x xdx

sin

sin sin(log )log

log 2

2 22

3

6+ −∫ is

(a)1

4

3

2log (b)

1

2

3

2log (c) log

3

2(d)

1

6

3

2log

Let ( , )x y0 0 be the solution of the following equations ( ) ( )log log2 32 3x y= ,

3 2log logx y= , then x 0 is equal to

(a)1

6(b)

1

2(c)

1

2(d) 6

Let P = − = : sin cos cos θ θ θ θ2 and Q = + = : sin cos sin θ θ θ θ2 be twosets. Then,(a) P Q⊂ and Q P− ≠ Φ (b) Q P⊄(c) P Q⊄ (d) P Q=

IIT JEE

UNSOLVED PAPER 2011

Mathematics

Let α and β be the roots of x x2 6 2 0− − = with α β> . If ann n= −α β for n ≥ 1,

then the value ofa a

a

10 8

9

2

2

−is

(a) 1 (b) 2 (c) 3 (d) 4

SECTION II


four choices (a), (b), (c) and (d) out of which one ore more may be correct.

Let M and N be two 3 3× non-singular skew-symmetric matrices such that

MN NM= . If P T denotes the transpose of P, then M N M N MNT T2 2 1 1( ) ( )− − is

equal to(a) M 2 (b) −N2 (c) −M 2 (d) MN

. Let the eccentricity of the hyperbolax

a

y

b

2

2

2

21− = be reciprocal to that of the

ellipse x y2 24 4+ = . If the hyperbola passes through a focus of the ellipse, then

(a) the equation of the hyperbola isx y2

2

2

23 21− =

(b) a focus of the hyperbola is ( , )2 0

(c) the eccentricity of the hyperbola is5

3

(d) the equation of the hyperbola is x y2 23 3− =

The vector(s) which is/are coplanar with vectors $ $ $i j k+ + 2 and $ $ $i j k+ +2 , are

perpendicular to the vector $ $ $i j k+ + is/are

(a) $ $j k− (b) − +$ $i j (c) $ $i j− (d) − +$ $j k

Let f R R: → be a function such that f x y f x f y( ) ( ) ( )+ = + , ∀ x, y R∈ . If f x( ) is

differentiable at x = 0, then

(a) f x( ) is differentiable only in a finite interval containing zero

(b) f x( ) is continuous, ∀ ∈x R

(c) f x′ ( ) is constant, ∀ ∈x R

(d) f x( ) is differentiable except at finitely many points

SECTION III

Directions (Q. Nos. 12-16) This section contains 2 paragraphs. Based upon one of the

paragraphs 2 multiple choice questions and based on the other paragraph 3 multiple choice

questions have the be answered. Each of these questions has four choices (a), (b), (c) and (d) out

of which only one is correct.

Paragraph (Q. Nos. 12-13)

IIT JEE (Mathematics) • Solved Paper 41

Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls and U2 contains only1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and putintoU2. However, if tail appears then 2 balls are drawn at random fromU1 and put intoU2. Now,1 ball is drawn at random from U2.

The probability of the drawn ball from U 2 being white is

(a)13

30(b)

23

30(c)

19

30(d)

11

30

Given that the drawn ball from U 2 is white, the probability that head appeared

on the coin is

(a)17

23(b)

11

23(c)

15

23(d)

12

23

Paragraph (Q. Nos. 14-16)

Let a b, and c be three real numbers satisfying [ ]a b c

1 9 7

8 2 7

7 3 7

= [ ]0 0 0 …(E)

If the point P a b c( , , ), with reference to Eq. (E), lies on the plane 2 1x y z+ + = ,

then the value of 7a b c+ + is

(a) 0 (b) 12 (c) 7 (d) 6

Let ω be a solution of x 3 1 0− = with Im ( )ω > 0. If a = 2, with b and c satisfying

Eq. (E), then the value of3 1 3

ω ω ωa b c+ +

(a) − 2 (b) 2 (c) 3 (d) − 3

Let b = 6, with a and c satisfying Eq. (E). If α and β are the roots of the quadratic

equation ax bx c2 0+ + = , then1 1

0 α β+

=

∞

∑n

n

is

(a) 6 (b) 7 (c)6

7(d) ∞

SECTION IV

Directions (Q. Nos. 17-23) This section contains 7 questions. The answer to each of the

questions is a single digit integer, ranging from 0 to 9.

The positive integer value of n > 3 satisfying the equation

1

sinπn

=

+

1

2

1

3sin sin

π πn n

is

Let a a a a1 2 3 100, , , ...., be an arithmetic progression with a1 3= and S ap i

i

p

==∑

1

,

1 100≤ ≤p . For any integer n with 1 20≤ ≤n , let m n= 5 . IfS

S

m

n

does not depend

on n, then a2 is

42 IIT JEE (Mathematics) • Solved Paper 2011

The minimum value of the sum of real numbers a a a a− − −5 4 3 83 1, , , , and a10

with a > 0 is

Let f : [ , ) [ , )1 2∞ → ∞ be a differentiable function such that f ( )1 2= . If

6 3 3

1f t dt x f x x

x( ) ( )= −∫ for all x ≥ 1, then the value of f ( )2 is

Let f ( ) sin tansin

cosθ θ

θ=

−1

2, where − < <π θ π

4 4. Then, the value of

d

df

(tan )( ( ))

θθ is

Consider the parabola y x2 8= . Let ∆1 be the area of the triangle formed by the

end points of its latusrectum and the point P1

22,

on the parabola and ∆ 2 be

the area of the triangle formed by drawing tangents at P and at the end points of

the latusrectum. Then,∆∆

1

2

is

If z is any complex number satisfying| |z i− − ≤3 2 2, then the maximum value

of| |2 6 5z i− + is

Paper II

SECTION I



Let P( , )6 3 be a point on the hyperbolax

a

y

b

2

2

2

21− = . If the normal at the point P

intersects the X-axis at (9, 0), then the eccentricity of the hyperbola is

(a)5

2(b)

3

2(c) 2 (d) 3

Let ( , )x y be any point on the parabola y x2 4= . Let P be the point that divides

the line segment from ( , )0 0 to ( , )x y in the ratio 1 3: . Then, the locus of P is

(a) x y2 = (b) y x2 2= (c) y x2 = (d) x y2 2=

Let f x x( ) = 2 and g x x( ) sin= for all x R∈ . Then, the set of all x satisfying

( )( ) ( )( )fogogof x gogof x= , where ( )( ) ( ( ))fog x f g x= is

(a) ± nπ, n ∈ , , , 0 1 2 K (b) ± ∈n nπ, , , ....1 2

(c)π π2

2+ n , n∈ − −..., , , , , , ...2 1 0 1 2 (d) 2 2 1 0 1 2n nπ, ..., , , , , , ...∈ − −


Let f : [ , ] [ , )− → ∞1 2 0 be a continuous function such that f x f x( ) ( )= −1 for all

x ∈ −[ , ]1 2 . Let R x f x dx1 1

2=

−∫ ( ) and R2 be the area of the region bounded by

y f x= ( ), x = −1, x = 2 and the X-axis. Then,

(a) R R1 22= (b) R R1 23= (c) 2 1 2R R= (d) 3 1 2R R=

If lim [ log( )] sin/

x

xx b b→

+ + =0

2 1 21 1 2 θb > 0and θ π π∈ −( , ], then the value of θ is

(a) ± π4

(b) ± π3

(c) ± π6

(d) ± π2

The circle passing through the point ( , )− 1 0 and touching the Y-axis at ( , )0 2 , also

passes through the point

(a) −

3

20, (b) −

5

22, (c) −

3

2

5

2, (d) ( , )− −1 4

Let ω ≠ 1 be a cube root of unity and S be the set of all non-singular matrices of

the form

1

1

1

a b

cωω ω2

, where each of a b, and c is either ω or ω2. Then, the

number of distinct matrices in the set S is

(a) 2 (b) 6 (c) 4 (d) 8

The value of b for which the equations x bx2 1 0+ − = , x x b2 0+ + = have

one root in common is

(a) − 2 (b) − i 3 (c) i 5 (d) 2

SECTION II



If f x

x

x

x

x

x

x

xx

( )

,

cos ,

,

log ,

=

− −

−

−

≤ −

− < ≤

< ≤>

π π

π2

1

2

20

0 11

, then

(a) f x( ) is continuous at x = − π2

(b) f x( ) is not differentiable at x = 0

(c) f x( ) is differentiable at x = 1 (d) f x( ) is differentiable at x = − 3

2

Let L be a normal to the parabola y x2 4= . If L passes through the point ( , )9 6 ,

then L is given by

(a) y x− + =3 0 (b) y x+ − =3 33 0 (c) y x+ − =15 0 (d) y x− + =2 12 0


Let E and F be two independent events. The probability that exactly one of themoccurs is 11/25 and the probability of none of them occurring is 2/25. If P T( )denotes the probability of occurrence of the event T, then

(a) P E P F( ) , ( )= =4

5

3

5(b) P E( ) = 1

5, P F( ) = 2

5

(c) P E P F( ) , ( )= =2

5

1

5(d) P E P F( ) , ( )= =3

5

4

5

Let f R: ( , )0 1 → be defined by f xb x

bx( ) = −

−1, where b is a constant such that

0 1< <b . Then,(a) f if not invertible on ( , )0 1

(b) f f≠ −1 on ( , )0 1 and f bf

′ =′

( )( )

1

0

(c) f f= −1 on ( , )0 1 and f bf

′ =′

( )( )

1

0

(d) f −1 is differentiable on (0, 1)

SECTION III

Directions (Q. Nos. 13-18) This section contains 6 questions. The answer to each of thequestions is a single digit integer, ranging from 0 to 9.

Let y x y x g x g x g x′ + ′ = ′( ) ( ) ( ) ( ) ( ) y( ) ,0 0= x R∈ , where f x′ ( ) denotesd f x

dx

( )and

g x( ) is a given non-constant differentiable function on R with g g( ) ( )0 2 0= = .Then, the value of y( )2 is...

Let a i k= − −$ $ , b i j= − +$ $ and c i j k= + +$ $ $2 3 be three given vectors. If r is avector such that r b c b× = × and r a⋅ = 0, then the value of r b⋅ is

Let ω π /= e i 3 and a b c x y z, , , , , be non-zero complex numbers such thata b c x+ + = , a b c y+ + =ω ω2 , a b c z+ + =ω ω2 .

Then, the value of| | | | | |

| | | | | |

x y z

a b c

2 2 2

2 2 2

+ ++ +

is

Let M be a 3 3× matrix satisfying

M

0

1

0

1

2

3

=−

, M

1

1

0

1

1

1

−

=−

and M

1

1

1

0

0

12

=

Then, the sum of the diagonal entries of M is

The number of distinct real roots of x x x x4 3 24 12 1 0− + + − = is

The straight line 2 3 1x y− = divides the circular region x y2 2 6+ ≤ into two

parts. If S =

−

2

3

4

5

2

3

4

1

4

1

4

1

8

1

4, , , , , , ,

, then the number of point(s) in S

lying inside the smaller part is


SECTION IV

Directions (Q. Nos. 19-20) This section contains 2 questions. Each question has fourstatements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II.Any given statement in Column I can have correct matching with one or more statement(s) givenin Column II.

Match the statements given in Column I with the values given in Column II.

Column I Column II

(A) If a = +$ $j k3 , b = − $ $j + k3 and c = 2 3 $k form a triangle,

then the internal angle of the triangle between a and b is(p)

π6

(B) If [ ( ) ]f x x dx a ba

b− = −∫ 3 2 2, then the value of f

π6

is (q)2

3

π

(C) The value ofπ π

2

7 6

5 6

3logsec( )x dx∫ is

(r) π /3

(D) The maximum value of arg1

1 −

zfor| | ,z z= ≠1 1 is given by (s) π

(t) π/2

Match the statements given in Column I with the intervals/union of intervalsgiven in Column II.

Column I Column II

(A) The set Re2

1 2

iz

z−

: z is a complex number,| |z = 1,

z ≠ ± 1

(p) ( , ) ( , )− ∞ − ∪ ∞1 1

(B) The domain of the function f xx

x( ) sin

( )( )

=−

−−

−1

2

2 1

8 3

1 3is (q) ( , ) ( , )− ∞ ∪ ∞0 0

(C) If f( )

tan

tan tan

tan

θθ

θ θθ

= −− −

1 1

1

1 1

, then the set

( ) : f θ θ π0

2≤ < is

(r) [ , )2 ∞

(D)If f x x x( ) ( )= −3 2 3 10 , x ≥ 0, then f x( ) is increasing in

(s) ( , ] [ , )− ∞ − ∪ ∞1 1

(t) ( , ] [ , )− ∞ ∪ ∞0 2

ANSWERS

Paper I1. (c) 2. (b) 3. (b) 4. (a) 5. (c) 6. (d)

7. (c) 8. (c) 9. (b, d) 10. (a, d) 11. (b, c) 12. (b)

13. (d) 14. (d) 15. (a) 16. (b) 17. (7) 18. (3 or 9)

19. (8) 20. (8/3) 21. (1) 22. (2) 23. (5)

Paper II1. (b) 2. (c) 3. (b) 4. (c) 5. (d) 6. (d)7. (a) 8. (b) 9. (a, b, c, d) 10. (a, b, d) 11. (a, d) 12. (a)

13. (0) 14. (9) 15. (3) 16. (9) 17. (2) 18. (2)19. (A) q; (B) p; (C) s; (D) s→ → → → 20. (A) s; (B) t; (C) r; (D) r→ → → →


Hints & Solutions

Paper I

1. Let v a b= + λ

v = + + − + +( ) $ ( ) $ ( ) $1 1 1λ λ λi j k

Projection of v on c = 1

3

⇒ v c

c

⋅ =| |

1

3

⇒ ( ) ( ) ( )1 1 1

3

1

3

+ − − − + =λ λ λ

⇒ 1 1 1 1+ − + − − =λ λ λ⇒ λ − =1 1 ⇒ λ = 2

∴ v = − +3 3$ $ $i j k

2. Here, area between 0 to b is R1 and b to

1 is R2.

∴ ( ) ( )1 11

42 21

0− − − =∫∫ x dx x dx

b

b

⇒ ( ) ( )1

3

1

3

1

4

3

0

3 1−−

− −

−

=x x

b

b

⇒ − − − + − −1

31 1

1

30 13 3( ) ( ) b b

= 1

4

⇒ − − = − +2

31

1

3

1

43( )b = − 1

12

⇒ ( )11

83− =b

⇒ ( )11

2− =b ⇒ b = 1

2

3. A straight line passing through P andmaking an angle of α = °60 , is given byy y

x x

−−

= ±1

1

tan ( )θ α

where, 3 1x y+ =⇒ y x= − +3 1

Then, tan θ = − 3

⇒ y

x

+−

= ±2

3 1

tan tan

tan tan

θ αθ αm

⇒ y

x

+−

= − +− −

2

3

3 3

1 3 3( )( )

andy

x

+−

= − −+ −

2

3

3 3

1 3 3( )( )

⇒ y + =2 0

andy

x

+−

= −−

=2

3

2 3

1 33

⇒ y x+ = −2 3 3 3

Neglecting, y + =2 0 as it does notintersect Y-axis.

4. Put x t2 = ⇒ x dxdt=2

∴ It

dt

t t=

⋅

+ −∫sin

sin sin(log )log

log 262

3…(i)

Using, f x dx f a b x dxa

b

a

b( ) ( )= + −∫∫

It

t= + −

+ − +−

1

2

2 3

2 3

6

sin(log log )

sin(log log ) sin

(log (log 2 3

2

3

+ −

∫log ))

log

log

t

dt

It

t tdt= −

− +∫1

2

6

62

3 sin(log )

sin(log ) sin( )log

log

∴ It

t tdt= −

− +∫sin (log )

sin (log ) sinlog

log 6

62

3…(ii)

On adding Eqs. (i) and (ii), we get

21

2

6

62

3I

t t

t tdt= + −

− +∫sin sin (log )

sin (log ) sinlog

log

∴ 21

223I t= ( )log

log = −1

23 2(log log )

⇒ I =

1

4

3

2log

5. Taking log on both sides,

log log ( ) log (log )2 2 3 3⋅ =x y

P(3, –2)

tan = 3a Ö

60°

60°

Ö3x + y = 1

⇒ log log log 2 2 + x

= +log log log 3 3 y …(i)

and log log log logx y⋅ =3 2

loglog log

logy

x= ⋅ 3

2…(ii)

From Eqs. (i) and (ii), we get

log log log 2 2 + x

= ⋅ + ⋅

log loglog log

log3 3

3

2

x

⇒ (log ) log log2 22 + ⋅ x

= + ⋅(log )(log )

(log )log3

3

22

2

x

⇒ log(log )

loglogx

3

22

2

−

= −(log ) (log )2 32 2

⇒ log(log ) (log )

logx

3 2

2

2 2−

= −(log ) (log )2 32 2

⇒ log log logx = − = −2 2 1

∴ x = 1

2

6. P = − = : sin cos cos θ θ θ θ2

⇒ cos ( ) sinθ θ2 1+ =⇒ tan θ = +2 1 …(i)

Q = + = : sin cos sin θ θ θ θ2

⇒ sin ( ) cosθ θ2 1− =

⇒ tan θ =−

× ++

1

2 1

2 1

2 1

= +( )2 1 …(ii)

∴ P Q=

7.a a

a

10 8

9

10 10 8 8

9

2

2

2

2

− = − − −−

( ) ( )

( )

α β α βα β9

= − − −−

α α β βα β

8 2 8 2

9 9

2 2

2

( ) ( )

( )

[Q α is root of x x2 6 2 0− − =α α2 2 6− = ]

Also, β is root of x x2 6 2 0− − =⇒ β β2 2 6− =

= −−

α α β βα β9

8 8

9

6 6

2

( ) ( )

( )= −

−=6

23

9 9

9 9

( )

( )

α βα β

8. Given, M M N NT T= − = −,

and MN NM= …(i)

∴ M N M N MNT T2 2 1 1( ) ( )− −

= ⋅− − −M N N M N MT T T2 2 1 1 1( ) ( )

= − −− − −M N NN M N MT2 1 1 1( ) ( ) ( ) ( )

= − − −− −M NI M N M2 1 1( )( ) ( )

= − − −M NM N M2 1 1

= − ⋅ − −M MN M N M( ) 1 1

= − − −M NM M N M( ) 1 1

= − − −MN NM N M( )1 1

= − −M NN M( )1 = −M 2

Note Here, non-singular word should notbe used, since there is no non-singular3 3× skew-symmetric matrix.

9. Here, equation of ellipse

x y2 2

4 11+ =

⇒ eb

a

22

21 1

1

4

3

4= − = − =

∴ e = 3

2and focus (± ae, 0)

= ±( , )3 0

For hyperbolax

a

y

b

2

2

2

21− = ,

eb

a12

2

21= + where, e

e12

2

1 4

3= =

⇒ 14

3

2

2+ =b

a⇒ b

a

2

2

1

3= …(i)

and hyperbola passes through ( , )± 3 0 .

Now,3

12a

= ⇒ a2 3= …(ii)


b2 1= …(iii)

∴ Equation of hyperbola is

x y2 2

3 11− =

Focus is ( , )± ae 0 .

Now, ± ⋅

32

30, ⇒ ( , )± 2 0

Hence, both options (b) and (d) are correct.

10. Let a b= + + = + +$ $ $ , $ $ $i j k i j k2 2

and c = + +$ $ $i j k


∴ A vector coplanar to a and b, andperpendicular to c.

Now, λ ( )a b c× ×⇒ λ ( ) ( ) a c b b c a⋅ − ⋅⇒ λ ( ) ($ $ $ )1 1 4 2+ + + +i j k

− + + + +( ) ($ $ $ )1 2 1 2i j k

⇒ λ $ $ $ $ $ $ 6 12 6 6 6 12i j k i j k+ + − − −

⇒ λ $ $ 6 6j k− ⇒ 6λ ($ $ )j k−

For λ = 1

6⇒ Option (a) is correct.

For λ = − 1

6⇒ Option (d) is correct.

11. f x y f x f y( ) ( ) ( ),+ = + as f x( ) is

differentiable at x = 0.

⇒ f k′ =( )0 …(i)

Now, ′ = + −→

f xf x h f x

hh( ) lim

( ) ( )

0

= + −→

lim( ) ( ) ( )

h

f x f h f x

h0

=→

lim( )

h

f h

h0

0

0from

Given, f x y f x f y x y( ) ( ) ( ), ,+ = + ∀∴ f f f( ) ( ) ( ),0 0 0= +when x y= = 0 ⇒ f( )0 0=Using L’Hospital’s rule,

lim( )

( )h

f hf k

→

′ = ′ =0 1

0 …(ii)

⇒ ′ =f x k( ) , on integrating both sides,

f x kx C( ) = + , as f( )0 0= ⇒ C = 0

So, f x kx( ) =∴ f x( ) is continuous for all x R∈ and

′ =f x k( ) , i.e. constant for all x R∈ .

Hence, both (b) and (c) are correct.

Solutions (Q. Nos. 12-13)

Head appears.

Tail appears.

12. Now, probability of the drawn ball fromV2 being white is

⇒ P (white/V2)

= ⋅ × + ×

P HC

C

C

C

C

C

C

C( )

31

51

21

21

21

51

11

21

+ × + ×

P TC

C

C

C

C

C

C

C( )

32

52

32

32

22

52

11

32

+ ⋅ ×

31

21

52

21

32

C C

C

C

C

Now, P (white/V2) = × + ×

1

2

3

51

2

5

1

2

+ × + × + ×

1

2

3

101

1

10

1

3

6

10

2

3= 23

30

13. P (head appeared/white from V2)

= ⋅× + ×

P H

C

C

C

C

C

C

C

C( )

31

51

21

21

21

51

11

21

23

30

=× + ×

1

2

3

51

2

5

1

223

30

= 12

23

Solutions (Q. Nos. 14-16)

Given, [ ] [ ]a b c 1 3

3 3

1 9 7

8 2 7

7 3 7

0 0 0×

×

=

⇒a b c

a b c

a b c

+ ++ ++ +

=

8 7

9 2 3

7 7 7

0

0

0

⇒ a b c+ + =8 7 0 …(i)

⇒ 9 2 3 0a b c+ + = …(ii)

⇒ a b c+ + = 0 …(iii)

IIT JEE (Mathematics) • Solved Paper 2011 49

3 W2 R

1 W

12

3

u1 u2

Initial

2 W2 R

2 W

12

3

u1 u2 2 cases

1W

3 W1 R

1 W1 R

u1 u2

1Ror

1 W2 R

3 W

12

3

u1 u2

3 cases

2W

3 W0 R

1 W2 R

u1 u2

2 W1 R

2 W1 R

u1 u2

On multiplying Eq. (iii) by 2, thensubtract from Eq. (ii), we get

7 0a c+ = …(iv)

Again multiplying Eq. (iii) by 3, thensubtract from Eq. (ii), we get

6 0a b− = …(v)

∴ b a= 6 and c a= −7

14. As ( , , )a b c lies on 2 1x y z+ + =⇒ 2 1a b c+ + =⇒ 2 6 7 1a a a+ − =⇒ a b c= = = −1 6 7, and

∴ 7a b c+ + = + − =7 6 7 6

15. If a = 2 , b = 12 and c = −14

∴ 3 1 3

ω ω ωa b c+ +

⇒ 3 1 32 12 14ω ω ω

+ + − = + +31 3

22

ωω

= + +3 1 3 2ω ω= + +1 3 2( )ω ω= − = −1 3 2

16. If b = 6 , a = 1 and c = −7

∴ ax bx c2 0+ + =⇒ x x2 6 7 0+ − =⇒ ( ) ( )x x+ − =7 1 0

∴ x = −1 7,

⇒ 1

1

1

70

−

=

∞

∑n

n

⇒ 6

70

=

∞

∑n

n

⇒ 16

7

6

7+ +

+ … ∞n

=−

1

16

7

= =1

1 77

/

17. Given, n > ∈3 Integer

and1 1

2

1

3sin sin sin

π π πn n n

=

+

⇒ 1 1

3

1

2sin sin sin

π π πn n n

− =

⇒sin sin

sin sin sin

3

3

1

2

π π

π π πn n

n n n

−

⋅=

⇒ 22

cos sinπ πn n

⋅ =⋅sin sin

sin

π π

πn n

n

3

2

⇒ 22 3

sin cos sinπ π πn n n

⋅ =2

⇒ sin sin4 3π πn n

=

⇒ 4 3π π πn n

= − ⇒ 7π πn

= ⇒ n = 7

18. Given, a m n1 3 5= =,

and a a1 2, ,… are in AP.

∴ S

S

S

S

m

n

n

n

= 5 is independent of n.

Now,

5

22 3 5 1

22 3 1

nn d

nn d

[ ( ) ]

[ ( ) ]

× + −

× + −

⇒ f d n

d n

( )

( )

6 5

6

− +− +

independent of n, if

6 0− =d ⇒ d = 6

∴ a a d2 1= + = + =3 6 9

Or

If d = 0S

S

m

n

is independent of n.

∴ a2 3=

19. Using AM ≥ GM,

a a a a a a a− − − − −+ + + + + + +5 4 3 3 3 8 101

8

≥ ( )a a a a a a a− − − − −⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅5 4 3 3 3 8 101

81

⇒ a a a a a− − −+ + + + +5 4 3 8 103 1 ≥ ⋅8 1

Hence, minimum value is 8.

20. Given, f( )11

3= and 6

1f t dt

x( )∫

= −3 3x f x x( ) , for all x ≥ 1

Using (Newton-Leibnitz formula),

On Differentiating both sides,

6 1 0 3f x f x( ) ( )⋅ − − + ′ −3 3 2x f x x( )

⇒ 3 3 3 2x f x f x x′ − =( ) ( )

⇒ f xx

f x x′ − =( ) ( )1


⇒ x f x f x

x

′ − =( ) ( )2

1 ⇒ d

dx

f x

x

( )

= 1

On integrating both sides,f x

xx C

( ) = + [Q f( )11

3= ]

⇒ 1

31= + C ⇒ C = − 2

3

Now, f x x x( ) = −2 2

3

⇒ f( )2 44

3= − = 8

3

Note Here, f( )1 2= does not satisfy given

function.

∴ f( )11

3=

For that, f x x x( ) = −2 2

3

and f( )2 44

3= − = 8

3

21. f( ) sin tansin

cosθ θ

θ=

−1

2, − < <π θ π

4 4

Let tansin

cos

− =1

2

θθ

φ

⇒ tansin

cosφ θ

θ=

2

∴ sinsin

sin cosφ θ

θ θ=

+2 2

=−

sin

sin

θ

θ1 2= =sin

costan

θθ

θ

∴ f( ) sin tanθ φ θ= =

⇒ d f

d

( )

(tan )

θθ

= 1

22. As, we know area of triangle formed bythree points on parabola is twice thearea of triangle formed bycorresponding tangents, i.e. area of∆PQR = 2 area of ∆T T T1 2 3.

∴ ∆ ∆1 22= or∆∆

1

2

2=

23. Given,| |z i− − ≤3 2 2 …(i)

To find minimum of| |2 6 5z i− +

or 2 35

2z i− +

Using triangle inequality,

i.e. z z z z1 2 1 2− ≤ +

∴ z i− +35

2

= − − + +z i i i3 2 25

2

= − − +( )z i i3 29

2≥ − − −z i3 2

9

2

≥ −29

2≥ 5

2⇒ z i− + ≥3

5

2

5

2

or 2 6 5 5z i− + ≥

Paper II1. Equation of normal to hyperbola at

( , )x y1 1 is

a x

x

b y

ya b

2

1

2

1

2 2+ = +

∴ At (6, 3),a x b y

a b2 2

2 2

6 3+ = +

It passes throught (9, 0).

Now,a

a b2

2 29

6

⋅ = +

⇒ 3

2

22 2a

a b− = ⇒ a

b

2

22=


√sin 2θ +cos2θ

√ θcos 2

sin θ

T2

T1 P

R

X

T3

Y

Q

∴ eb

a

22

21= + = +1

1

2⇒ e = 3

2

2. By section formula,

hx= + 0

4, k

y= + 0

4

∴ x h y k= =4 4, and

Substituting in y x2 4= ,

( ) ( )4 4 42k h= ⇒ k h2 =or y x2 = is required locus.

3. f x x g x x( ) , ( ) sin= =2

( ) ( ) singof x x= 2

go gof x x( ) ( ) sin(sin )= 2

( ) ( ) (sin(sin ))fogogof x x= 2 2 …(i)

Again, ( ) ( ) singof x x= 2

( ) ( ) sin(sin )gogof x x= 2 …(ii)

Given, ( ) ( ) ( ) ( )fogogof x gogof x=⇒ (sin(sin )) sin(sin )x x2 2 2=⇒ sin(sin ) sin(sin ) x x2 2 1 0− =⇒ sin(sin )x2 0= or sin(sin )x2 1=

⇒ sin x2 0= or sin x2

2= π

∴ x n2 = π(i.e. not possible as − ≤ ≤1 1sin )θ

x n= ± π

4. Here, R x f x dx11

2=

−∫ ( ) …(i)

Using, f x dxa

b( )∫ = + −∫ f a b x dx

a

b( )

R x f x dx11

21 1= − −

−∫ ( ) ( ) ,

[Q f x f x( ) ( )= −1 ]

∴ R x f x dx11

21= −

−∫ ( ) ( ) …(ii)

Given, R2 is area bounded by

f x x( ), − 1 and x = 2

∴ R f x dx21

2=

−∫ ( ) …(iii)

Adding Eqs. (i) and (ii), we get

2 11

2R f x dx=

−∫ ( ) …(iv)

∴ From Eqs. (iii) and (iv), we get

2 1 2R R=

5. Here, lim log( ) /

x

xx b→

+ +0

2 11 1

[1∞ from]

⇒ exx b

xlim log( )→

+ ⋅0

211

⇒ e bblog( ) ( )1 22

1+ = + …(i)

Given,

lim log( ) sin/

x

xx b b→

+ + =0

2 1 21 1 2 θ

⇒ ( ) sin1 22 2+ =b b θ

∴ sin221

2θ = + b

b…(ii)

By AM ≥ GM,

bb b

b

+≥ ⋅

1

2

11 2/

⇒ b

b

2 1

21

+ ≥ …(iii)

From Eqs. (ii) and (iii), we get sin2 1θ =

⇒ θ π= ±2

as θ π π∈ −( , ]

6. Equation of circle passing through apoint ( , )x y1 1 and touching the straightline L, is given by

( ) ( )x x y y L− + − = =12

12 0λ

∴ Equation of circle passing through(0, 2) and touching x = 0.

Now, ( ) ( )x y x− + − + =0 2 02 2 λ …(i)

Also, it passes through ( , )− 1 0 .

So, 1 4 0+ − =λ ⇒ λ = 5

Eq. (i) becomes,

x y y x2 2 4 4 5 0+ − + + =⇒ x y x y2 2 5 4 4 0+ + − + =For x-intercept, put y = 0,

x x2 5 4 0+ + =⇒ ( ) ( )x x+ + =1 4 0

∴ x = − −1 4,

7. A /= 0, as non-singular.

∴1

1

1

02

a b

cωω ω

≠

⇒ 1 1 2( ) ( )− − −c a cω ω ω


y x2=4

X

Y

(0, 0)O

( ,x y)Q

3

1P h k( , )

+ − ≠b( )ω ω2 2 0

⇒ 1 02− − + ≠c a acω ω ω⇒ ( ) ( )1 1 0− − ≠c aω ω

⇒ a c≠ ≠1 1

ω ω, ⇒ a c= =ω ω,

and b ∈ , ω ω2 ⇒ 2 solutions

8. If a x b x c12

1 1 0+ + =and a x b x c2

22 2 0+ + =

have a common real root, then

⇒ ( ) ( )a c a c b c b c1 2 2 12

1 2 2 1− = −( )a b a b1 2 2 1−

∴ x bx

x x b

2

2

1 0

0

+ − =+ + =

have a common root.

⇒ ( ) ( )( )1 1 12 2+ = + −b b b

⇒ b b b b b2 2 32 1 1+ + = − + −⇒ b b3 3 0+ = ⇒ b b( )2 3 0+ =⇒ b i= ±0 3,

9. f x

x x

x x

x x

x x

( )

,

cos ,

,

log ,

=

− − ≤ −

− − < ≤

− < ≤>

π π

π2 2

20

1 0 1

1

Continuity at x = − π2

,

f −

= − −

− =π π π2 2 2

0

RHL ⇒ lim cosh

h→

− − +

=0 2

0π

∴ Continuous at x = 0

Continuity at x = 0 ⇒ f( )0 1= −RHL ⇒ h h→ + − = −0 0 1 1lim ( )


Continuity at x = 1; f( )1 0=RHL ⇒ lim log ( )

hh

→+ =

01 0


Here, f x

x

x x

x

xx

′ =

− ≤ −

− < ≤

< ≤

>

( )

,

sin ,

,

,

12

20

1 0 11

1

π

π

Differentiable at x = 0,

LHD = 0, RHD = 1

∴ Not differentiable at x = 0

Differentiable at x = 1,

LHD = 1, RHD = 1

∴ Differentiable at x = 1

Also, for x = − 3

2⇒ f x x( ) = − − 3

2

∴ Differentiable at x = − 3

2

10. Normal to y x2 4= , is

y mx m m− − −2 3 which passesthrough (9, 6).

Now, 6 9 2 3= − −m m m

⇒ m m3 7 6 0− + = ⇒ m = −1 2 3, ,

∴ Equation of normals are

y x− + =3 0,

y x+ − =3 33 0 and y x− + =2 12 0

11.

P E F P E F( ) ( )∪ − ∩ = 11

25…(i)

(i.e. only E or only F)

Neither of them occurs = 2

25

⇒ P E F( )∩ = 2

25…(ii)

From Eq. (i), we get

P E P F P E F( ) ( ) ( )+ − ∩ =211

25…(iii)

From Eq. (ii), we get

( ( )) ( ( ))1 12

25− − =P E P F

⇒ 12

25− − + ⋅ =P E P F P E P F( ) ( ) ( ) ( )

…(iv)

From Eqs. (iii) and (iv), we get

P E P F( ) ( )+ = 7

5and P E P F( ) ( )⋅ = 12

25


E F

E F

∴ P E P E( ) ( )⋅ −

=7

5

12

25

⇒ ( ( )) ( )P E P E2 7

5

12

250− + =

⇒ P E P E( ) ( )−

−

=3

5

4

50

∴ P E( ) = 3

4or

4

5⇒ P F( ) = 4

5or

3

5

12. Here, f xb x

bx( ) = −

−1

where, 0 1 0 1< < < <b x,

For function to be invertible it shouldbe one-one onto.

∴ Check range :

Let f x y( ) = ⇒ yb x

bx= −

−1

⇒ y bxy b x− = − ⇒ x by b y( )1 − = −

⇒ xb y

by= −

−1

where, 0 < x < 1

∴ 01

1< −−

<b y

by

b y

by

−−

>1

0 andb y

by

−−

<1

1

⇒ y b< or yb

> 1…(i)

( ) ( )b y

byy

b

− +−

< − < <1 1

11

1…(ii)


yb

∈ −

⊂11

, Codomain

Thus, f x( ) is not invertible.

13.dy

dxy g x g x g x+ ⋅ ′ = ′( ) ( ) ( )

IF = ∫ =′e e

g x dx g x( ) ( )

∴ Solution is

y e g x g x e dx Cg x g x( ) ( ) ( )( ) ( )= ⋅ ′ ⋅ +∫Put g x t g x dx dt( ) , ( )= ′ =

y e t e dt Cg x t( )( ) = ⋅ +∫= ⋅ − ⋅ +∫t e e dt Ct t1 = ⋅ − +t e e Ct t

y e g x e Cg x g x( ) ( )( ( ) )= − +1 …(i)

Given, y( )0 0= , g g( ) ( )0 2 0= =∴ Eq. (i) becomes,

y e g e Cg g( ) ( ( ) )( ) ( )0 0 10 0⋅ = − ⋅ +⇒ 0 1 1= − ⋅ +( ) C ⇒ C = 1

∴ y x e g x eg x g x( ) ( ( ) )( ) ( )⋅ = − +1 1

⇒ y e g eg g( ) ( ( ) )( ) ( )2 2 1 12 2⋅ = − +where, g( )2 0=⇒ y( ) ( )2 1 1 1 1⋅ = − ⋅ + ⇒ y( )2 0=

14. r b c b× = ×⇒ ( )r c b− × = 0 ⇒ r c b− + λor r c b= + λ …(i)

Given, r a⋅ = 0, taking dot product witha for Eq. (i).

Now, r a a c a b⋅ = ⋅ + ⋅λ

∴ λ = − ⋅⋅

r r

r ra c

a b[Q

r rr a⋅ = 0] …(i)


r ca c

a bb= − ⋅

⋅Taking dot with b, we get

r b c ba c

a bb b⋅ = ⋅ − ⋅

⋅⋅( )

where,

a

b

c

= − −= − += + +

$ $

$ $

$ $ $

i k

i j

i j k2 3

= − + − − − +( )( )

( )( )1 2

1 3

11 1 = + =1 8 9

15. Here, ω π /= ei2 3, then only integer

solution exists.

Then,x y z

a b c

2 2 2

2 2 23

+ +

+ +=

16. Let M

a a a

b b b

c c c

=

1 2 3

1 2 3

1 2 3

∴ M M

0

1

0

1

2

3

1

1

0

1

1

1

=−

−

=−

,

,


+

b 1/b

– +

M

1

1

1

0

0

12

=

⇒a

b

c

a a

b b

c c

2

2

2

1 2

1 2

1 2

1

2

3

=−

−−−

,

=−

1

1

1

,

a a a

b b b

c c c

1 2 3

1 2 3

1 2 3

0

0

12

+ ++ ++ +

=

⇒ a b c a a2 2 2 1 21 2 3 1= − = = − =, , , ,

b b c c1 2 1 21 1− = − = −,

⇒ a a a b b b1 2 3 1 2 30 0+ + = + + =, ,

c c c1 2 3 12+ + =∴ a b c1 2 30 2 7= = =, and

Hence, sum of diagonal elements

= + + =0 2 7 9

17. f x x x x x( ) = − + + −4 3 24 12 1

f x x x x′ = − + +( ) 4 12 24 13 2

f x x x′ ′ = − +( ) 12 24 242

= − +12 2 22( )x x

= − + >12 1 1 02( ) x , for all x

⇒ f x′ ( ) is increasing.

Since, f x′ ( ) is cubic and increasing.

⇒ f x′ ( ) has only one real root and twoimaginary roots.

∴ f x( ) cannot have all distinct root.

⇒ Atmost 2 real roots.

Now, f( )− =1 15,

f( )0 1= − and f( )1 9=∴ f x( ) must have one root in ( , )− 1 0and other in ( , )0 1 .

⇒ 2 real roots.

18. x y2 2 6+ ≤ and 2 3 1x y− = is shown as,

For the point to lie in the shaded part,origin and the point lie on opposite sideof straight line L.

∴ For any point in shaded part L > 0and for any point inside the circle S < 0.

Now, for 23

4,

L x y: 2 3 1− −

L : 49

41

3

40− − = >

and S x y: 2 2 6+ − ,

S : 49

166 0+ − <

⇒ 23

4,

lies in shaded part.

For5

2

3

4,

L : 5 9 1 0− − < [neglect]

For1

4

1

4

1

2

3

41 0, :−

+ − >

L

∴ 1

4

1

4, −

lies in the shaded part.

For1

8

1

4

1

4

3

41 0, :

− − <L [neglect]

⇒ Only 2 points lie in the shaded part.

19. (A) ∴ a = + =1 3 2

b = + =1 3 2

c = =12 2 3

Using cosine law,

cos C =+ −a b c

a b

2 2 2

2

= + −× ×

4 4 12

2 2 2= − 4

8= −1

2

⇒ ∠ = ° =C 1202

3

π


1—2

1—3

L

C

A B

b j k= – + 3Ö^ ^a j k= + 3Ö^ ^

c k= 2 3Ö ^

(B) f x dxx

a ba

b

a

b

( ) ( )−

= −∫ 3

2

22 2

⇒ f x dx b a a ba

b( ) ( ) ( )− − = −∫

3

22 2 2 2

⇒ f x dx a b b aa

b( ) ( ) ( )∫ = − + −2 2 2 23

2

= −b a2 2

2

⇒ f x dxb a

a

b( )∫ = −2 2

2

f x x( ) = ⇒ fπ π6 6

=

(C)π π

2

3 7 6

5 6

logsec( )

/

/

e

x dx∫

⇒ π π ππ

2

7 6

5 6

3log

log tan

/

/

e

x xsec +

⇒ π 5π6

5πlog

log tan3 6

sec +

− +

log tansec7π6

7π6

⇒ πlog

log log3

31

3−

⇒ π πlog

log 3

3 =

(D) arg1

1( )− z, for z =1

⇒ arg ( )1 1− −z

⇒ − −arg ( )1 z ⇒ arg ( )1 − z

From figure, arg ( )z − 1 is maximum = π.

20. (A) Given, z = 1 ⇒ z z⋅ = 1

∴ 2

1

2 22 2

iz

z

iz

z z z

i

z z−=

⋅ −=

−,

Let z x iy= +

∴ z z iy− = 2 =−

= −2

2

1i

iy y…(i)

where, y x= −1 2

∴ − ≤ ≤1 1y ⇒ − ≤1 y

and y ≤ 1 ⇒ − ≥11

yand

11

y≥

⇒ Re2

11 1

2

iz

z−

∈ − ∞ − ∪ ∞( , ] [ , )

(B) f xx

x( ) sin

( )( )

=−

−

−

−1

2

2 1

8 3

1 3,

For domain, − ≤−

≤−

−18 3

1 31

2

2 1

( )( )

x

x

⇒ − ≤ ⋅ −−

≤− −

−19 3 3

1 31

2 2

2 1

( ) ( )( )

x x

x

∴ − ≤ −− ⋅

≤−

−13 3

1 3 31

2

2

x x

x x( )

3 3

1 3 31

2

2

x x

x x

−− ⋅

≥ −−

−

⇒ ( ) ( )

( ) ( )

3 1 3 1

3 1 3 10

2

1 1

x x

x x

− −+ −

≥−

− −

⇒ x ∈ − ∞ ∪ ∞( , ] ( , )0 1

and3 3

1 3 31

2

2

x x

x x

−− ⋅

≤−

−

⇒ ( ) ( )

( ) ( )

3 1 3 1

3 1 3 10

2

1 1

x x

x x

−

− −− +

+ −≥

and x ∈ − ∞ ∪ ∞( , ) [ , )1 2

∴ x ∈ − ∞ ∪ ∞( , ] [ , )0 2

(C) f( )

tan

tan tan

tan

θθ

θ θθ

= −− −

1 1

1

1 1

R R R1 1 3→ + ,

f( ) tan tan

tan

θ θ θθ

= −− −

0 0 2

1

1 1

= +2 12(tan )θ = ≥2 22sec θf( ) [ , )θ ∈ ∞2


z

z

z

z

z1

–1 1

–1

0 1

1 2

physics - m.media-amazon.com...9 8 1 rt (b) 3 2 1 rt (c) 15 8 1 rt (d) 9 2 1 rt ... an electron and...

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