test rider advanced - m.media-amazon.com

PRACTICE SET 2 75

A to the Various IITs...S ingle Door Ent r y

Advanced

Paper 1

Duration : 3 Hours Max. Marks : 180

Test RIDER

Pra

cti

ce

Se

t2

Question Booklet Code 1

Please read the instructions carefully. You are allotted 5 minutes speciallyfor this purpose.

Question Paper Format

Marking Scheme

This booklet is your question paper. Attempt all the questions.

Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and

electronic gadgets are not allowed.

Write your name and roll number in the space provided on the bottom of this page.

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists

of three sections.

contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which only one is correct.

contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which one or more than one is/are correct.

contains 5 questions. The answer to each question is a single–digit integer, ranging

from 0 to 9 (both inclusive)

For each question in Section 1, you will be awarded for correct answer and zero mark for

unattempted questions. No negative marks will be awarded for incorrect answers in this section.

For each question in Section 2, you will be awarded for correct answer(s) and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

For each question in Section 3, you will be awarded for the correct answer and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

Section 1

Section 2

Section 3

2 marks

4 marks

4 marks

1. A conducting rod is rotating in a uniform magnetic perpendicular to it as shown in the figure.

If the rotation is now become clockwise with the same angular speed then what will be the

change in the potential difference between O and AP?

a. −B lω 2b. B lω2 2

c. 2 2 2B lω d. 3 2B lω

2. A uniform sphere of mass m and radius r is planed in a corner with initial

counter clockwise angular velocity ω0 as given. If µ =1 for both surfaces,

then time required for sphere to come to rest is

a.ω0 r

gb.

ω0

5

r

gc.

2 0ω r

gd.

2

50ω r

g

3. A uniform circular disc placed on a rough horizontal surface has

initially a velocity v 0 and angular velocity ω0as shown. Disc comes to

rest after moving certain distance, thenv

r

0

0ωwill be

a.1

2b. 1 c.

3

2d. 2

4. Two long straight wires with the same cross-section are arranged in air

parallel to one another. The distance between the axis of the wire is ηtimes larger than the radius of wires cross-section. The capacitance of

the wires per unit length would be [Take, η >> 1]

a.2 0πε

ηInb.

πεη

0

2 In

c.πε

η0

Inc. Information is insufficient

5. A boy with mass m is standing on the edge of the disc with moment of inertia I, radius R and

initial angular velocity ω . Boy jumps off the disc with tangential speed v with respect to ground.

The new angular velocity of disc will be

a.I mv

I

ω2 2−b.

(I mR mv

I

+2 2 2)ω −

c.I mvR

I

ω −d.

( )I+mR mvR

I

2 ω −

AdvancedTest RIDER

v0

ω0

Top view of thearrangement

ω0

Part I

Section 1 Single Correct Option Type

10

x x x x x x

x x x x x x

x x x x x x

x x x x x x

l

AO

ω

6. When a thin transparent sheet of refractive index µ =3

2is placed near one of the slit in YDSE, the

intensity at centre of screen reduces to half of maximum intensity. Then, minimum thickness of

sheet should be

a.λ4

b.λ8

c.λ2

d.λ3

7. A small particle of mass m and its retaining cord are spinning with angular

velocity ω on the horizontal surface of a smooth disc. As force F is slightly

relaxed, r increases and ω changes. Determine the rate of change of ω with

respect to r

a. + ωr

b. − ωr

c. − 2ωr

d. + 2ωr

8. Consider the combination of two discs, then what will be the velocity of falling disc centre as a

function of h. Both discs are identical and string does not slip relative to disc

a.8

5

ghb.

6

5

ghc.

gh

5d. gh

9. A combination of concave lens and concave mirror is placed with the common principal axis.

The rays coming from the infinity parallel to principal axis incident on the concave lens and final

image is formed at the infinity. Choose the correct options.

a. | | | |f f1 2= b. | | | |f f1 22< c. | | | |f f1 22> d. Information is sufficient

10. A homogeneous cylinder of massm and radius r is at rest on horizontal

plane when a couple is applied as shown. Determine magnitude of

coefficient of friction between the wheel and plane so that rolling

occur.

a. µ τ>mgR

b. µ τ> 3

2mgR

c. µ τ> 3

mgRd. µ τ> 2

3mgR

PRACTICE SET 2 77

rO

r

h

C

τR

Rough

F

ωrm

f2f1

11. Two blocks are connected by a ideal string which is initially slacked and one block is projected

along surface. Identify correct statements about subsequent motion

a. initial speed of CM is 2 3v/

b. speed of CM will remain same after string get taught

c. impulse due to tension on block of mass m will increase its momentum

d. impulse due to tension on block will decrease momentum of block of mass 2 m

12. In a given system m m1 2> , system is held at rest by thread BC. Just after the

thread BC is burnt

a. initial acceleration of m2 will be upwards

b. magnitude of acceleration of blocks ism m

m mg1 2

1 2

−+

c. initial acceleration of m1 will be zero

d. magnitude of acceleration of block is non-zero and unequal

13. A box is placed on the top most point of circular path as shown

in the figure. It is given a push and it starts sliding downwards.

Choose the correct options

a. It will leave the circular path at a point where N = 0

b. It will leave the circular path just after the instant when N = 0

c. When it leave the circular path the contact force at that instant on the circular path ismv

R

2

, where

v is the speed of block at that instant

d. When N = 0, there contact force on the circular path ismv

R

2

, where v is the speed at that instant.

14. In the circuit shown in figure, E1 and E2are two ideal sources of unknown emfs. Some currents

are shown. Potential difference appearing across 6 Ω resistance is V VA B− =10 V,

a. the current in the 4 Ω resistor in between C and D is 5 A

b. the unknown emf E1 is 36 V

c. the unknown emf E2 is 54 V

d. the resistance R is equal to 9 Ω

AdvancedTest RIDER

A

Bk

m1

m2

C

m 2m v

Smooth horizontal surface

FC

θ

mgcosθmg sin θ

mg

N

θv

2 A

4 Ω 3 Ω 3 Ω 6 Ω

4 ΩC D B

4 A

A

E1 E2

R

Section 2 More Than One Correct Option

This section contains . Each question has four choices, (a), (b), (c)

and (d) out of which or is/are correct.

5 multiple choice questions

one more than one

15. A photon moves vertically up in a region with gravitational field g downwards. The frequency of

photon at an instant is ν. After it has moved up by height h,

a. its speed decreases

b. its energy decreases

c. its frequency is ν 12

−

gh

c

d. its frequency is νe

gh

c

−2

16. A hydrogen atom is in its ground state and stationary, when a neutron of kinetic energy 25.5 eV

collides with it. Take mass of hydrogen atom = mass of neutron = × −1.6 10 27 kg and

1eV 1.6 10 19= × − J, if the collision is head-on, find the maximum possible number of photons of

different wavelength that can be emitted after the collision from the hydrogen atom.

17. Let us consider that the value of gravitational potential on the surface of earth to be zero. A

particle was initially at point at distance Re from the earth’s surface. The work done to bring this

particle to earth’s surface is−

XGM me

2Re

, so the value of X is? (m = mass of particle)

18. Consider a horizontal surface moving vertically upward with velocity 2 m/s. A small ball of mass

2 kg is moving with velocity 2 2$ $i j− (m/s). If coefficient of restitution is 1/2 and friction

coefficient is 1/3. Find horizontal component of velocity of ball after collision

19. A certain series R-C circuit is formed using a resistance R, a capacitor without

dielectric having a capacitance C =2 F and a battery of emf E =3 V. The

circuit is completed and it is allowed to attain the steady state. After this, at

t =0, half the thickness of the capacitor is filled with a dielectric constant

K =2 as shown in the figure. The system is again allowed to attain a steady

state. What will be the heat generated (in joule) in the capacitor between t =0

and t = ∞?

20. A uniform rope of length L and massm is held at one end and whirled in a horizontal circle with

angular velocity ω. You can ignore the force of gravity on rope. The time required for a

transverse wave to travel from one end of rope to other isπ

ωn, then the value of n is

PRACTICE SET 2 79

y

x

2 m/s

Section 3 Integer Answer Type

This section contains . The answer to each question is a single-digit

integer, ranging from to (both inclusive).


0 9

K = 2

d/2

d

Part II

21. 92238U by successive decay changes to 82

206U , when a sample of uranium ore was analysed it

was found that it contains 1g of U238 and 0.1g of Pb206 , considering that all the Pb206 had

accumulated due to decay of U238 . Calculate the age of the ore. (Half-life of U238 945 10= ×. yrs).

a. 01155 108. × yrs b. 7 099 108. × yrs

c. 0154 10 9. × − yrs d. 7 099 1010. × yrs

22. Given that, λ 0 and λ are threshold wavelength and wavelength of incident light respecturely

then what will be the velocity of ejected electron from the metal surface?

a.2

0

h

m( )λ λ− b.

20

hc

m( )λ λ− c.

2 0hc

m

λ λλλ 0

−

d.

2 1 1

0

h

m λ λ−

23. Which of the following is the suitable monomer for cationic polymerisation?

a. CH CH C

O

OMe2

⊕

b. CH C

H

O C

O

CH2 3

⊕

c. CH CH Cl2 ⊕

d. CH CH CN2 ⊕

24. Which of the following pairs of complexes are isomeric with each other but their aqueous

solutions exhibit different molar conductivities?

a. [PtCl (NH ) ]Br2 3 4 2 and [PtBr (NH ) ]Cl2 3 4 2

b. [CoCl (NH ) ]NO2 3 4 2 and [CoCl(NO )(NH ) ]Cl2 3 4

c. [Co(NO )(NH ) ]Cl2 3 5 2 and [Co(ONO)(NH ) ]Cl3 5 2

d. [CoBr (NH ) ]SO2 3 4 4 and [Co(SO )(NH ) ]Br4 3 4 2

25. The efficiency of a reversible cycle is shown as in figure is

a. 33.33% b. 52.3% c. 10% d. 29.2%

AdvancedTest RIDER


10

B

CA

200 K

100 K

T

500 1000

S (J/K)

26. Which of following is a method for preparation of Engel’s sulphur?

a. S Cl H S2 2 2 4+ b. 2HNO + H S3 2 c. 2H S + SO2 2 d. All of these

27. Phosphorus on reaction with hot solution of NaOH produces, A, which on reaction with nitric

acid produces B. B can also be obtained by direct oxidation of phosphorus in air, which of the

following is correct statement regarding A and B ?

a. A is, hydrate of phosphorus, B is oxide of phosphorus

b. A is, hydride of phosphorus, B is oxide of phosphorus

c. A is, oxide of phosphorus, B is hydride of phosphorus

d. A is, hydroxide of phosphorus, B is oxide of phosphorus

28.

a. b. c. d.

29. The standard emf of S2−/CuS/ Cu is, if K sp (CuS)= K and E 0 2cu cu+ / = V volt at 25° C.

a. V K+ 0 059

2

.log b. V K− 0 059

2

.log

c. 2 V K+ log d. V K2 0 059

2− .

log

30. Which of the following is an oxidation reaction?

a. CH CH CH3 24

2

CCl

Br == → b. CH CH CH

OH3 2

4KMnO == →

s

c. CH CH CH3 2 == →⊕H / H O2

d. Both a and b

31. Arrange the following inorganic compounds ions in their increasing orders of % s character.

I. CO 32− II. XeF 4 III. I 3

− IV. NCl 3 V. BeCl 2

a. II < III < IV < I < V b. II < IV < III < V < I

c. III < II < I < V < IV d. II < IV < III < I < V

32. Nitrogen forms more than one type of oxide which may or may not contain H—N bond. The wide

range of these types of compound is due to possibility of N to extend its oxidation upto 5. Which

of the following oxides of nitrogen will contain atleast one N N bond?

a. N O2 4 b. N O2 5 c. N O2 3 d. NO 2

PRACTICE SET 2 81

OH

H⊕P, P will be

OO





one more than one

33. Choose correctly matched columns.

A B

a. MeOH > EtOH > Me2CHOH > Me3COH Rate of esterification

b. Me3COH > Me2CHOH > MeCH2OH Rate of dehydration

c. HCOOH > MeCO2H > EtCO2H > Me3C CO2H Rate of esterification

d. MeCOCl >(MeCO)2O> MeCO2 Me > MeCONH2 Rate of reactivity towards CH3MgX

34. A compound X on treatment with sodium hexanitro cobaltate (III) produces yellow coloured

precipitate and on reaction with hexachloroplatinum hydride also produces yellow coloured

precipitate. Here X is

a. NH4+

b. Na+c. Mg2+

d. K+

35. Amino acid contains an acidic group —COOH as well as a basic group —NH 3 the structure of

alanine is NH CH

CH

COO3

3

⊕ −

, which of the following statements, is/are incorrect statement

about it?

a. It is an acidic amino acid

b. It is an essential amino acid

c. It will move to cathode in electrolysis at pH = 5

d. It is an optically active amino acid

36. Considering Z as internuclear axis. Find the number of delta bonding molecular orbital among

following set?

d d d dx y xy x y xz2 2 2 2− −

and , or

37. If the K sp of Mg(OH)2 is 1 10 12× − , then 0.01 M MgCl2 will precipitate at a limiting pH of.

38. Find the number of X M X− − angles in MX 6 , which has an octahedral geometry. If it is x thenx

2

equals

39. Two vessels divided by a partition contain 1 mol of N2 and 2 mol of O2 gas, If the partition is

removed and gases are mixed isothermally, then find the change in entropy due to mixing

assuming initial and final pressure are same. The answer after dividing by a factor of 2 comes as.

40. In the Lindemann theory of unimolecular reactions, it is shown that apparent rate constant for

such a reaction is kk c

capp =

+1

1 α, where c is the concentration of the reactant, k1 and α are

constants. Calculate the value of c for which k app has 90% of its limiting value at c tending to

infinitely large values, given α = ×9 105 . If answer is x then | log |x equals to

82 AdvancedTest RIDER





0 9

Part III

41. If α β, be the roots of x x2 1 0− − = and An = +α βn n , then AM of An−1 and An is

a.1

21An + b. 2 2An − c. 2 An +1 d. None of these

42. If sin −1 ( ) cos ( ) tan cosx xx

xk− + − +

−

= +− − −1 32

1 1

2

1 π, then the value of k is

a. 1 b. − 1

2c.

1

2d. None of these

43. Let there be two points A and B on the curve y x= 2 in the plane OXY satisfying OA i =⋅ $ 2 and

OB i⋅ =$ –3, then the length of the vector | − |3 4O A OB is

a. 30 b. 18 c. 24 d. 42

44. The number of solution of the equation | | =−∫ cos ,x dxx

02

02

< <xπ

will be

a. 0 b. 1 c. 2 d. 3

45. If f x( ) is an even function and satisfies the relation x f x fx

g x2 21

( ) ( )−

= , where g x( ) is an

odd function, then f ( )5 equals

a. 0 b.50

75c.

49

75d. None of these

46. If z is a complex number lying in the first quadrant such that Re ( ) Im( )z z+ =3, then the

maximum value of Re( ) Im( )z z2 is

a. 1 b. 2 c. 3 d. 4

47. Letdf x

dx

e

xx

x( ),

cos

= >0 . If3

1

3

1

5e

xdx f k f

xcos

( ) ( )∫ = − , then one of the possible values of k is

a. 122 b. 123 c. 124 d. 125

48. If a is an integer lying in [– , ]5 30 , then the probability that the graph of

y x a= + +2 2 4( ) x a− +5 64 is strictly above the x-axis is

a.7

36b.

1

6

c.2

9d.

3

5


10

49. The equation of the straight line in the plane r n⋅ =d , which is parallel to r a b= + λ and passes

through the foot of perpendicular drawn from the point P ( )a to the plane r n⋅ =d is

(where, n b⋅ =0)

a. r aa n

nn b= + − ⋅

+d λ b. r aa n

nn b= + − ⋅

| |

+d

2λ

c. r aa n

nn b= + ⋅ −

+d2

λ d. r aa n

nn b= + ⋅ −

+d λ

50. lim min ( )sin

xy y

x

x→− + ⋅

0

2 4 11 , (where [. ] denotes the greatest integer function) is

a. 5 b. 6 c. 7 d. does not exist

51. Given, x

f x

x→ =0 2

2lim ( ), where [⋅] denotes greatest integer function, then

a. lim [ ( )]x

f x→

=0

0 b. lim [ ( )]x

f x→

=0

1

c. lim( )

x

f x

x→

0

does not exists d. lim( )

x

f x

x→

0

exists

52. Let E1 and E2 be two ellipsesx

ay

2

2

2 1+ = and xy

a

22

21+ = (where a is a parameter). Then, the

locus of the points of intersection of the ellipses E1 and E2 is a set of curves comprising

a. two straight lines b. one straight lines

c. one circle d. one parabola

53. The sum to n terms of the series tan tan tan tan− − − −

+

+

+1 1 1 11

2

2

9

1

8

2

25

+

+…−tan 1 1

8n

terms

a. tan−

1 1

3b. tan ( )−1 3 c. cot−

1 1

3d. None of these

54. p and q are non-zero constants. The equation x px q2 0+ + = has roots u and v, then

a. q x px2 1 0+ + = has roots1 1

u v, b. ( )( )x − − =p x q 0 has roots u v+ and u v.

c. x p q x q2 2 22 0− − + =( ) has roots u2 and v2d. x qx p2 0+ + = has roots

u

vand

v

u

55. Let α β= + + = + +a b c b c a$ $ $ $ $ $i j k, i j k and γ = + +c a b$ $ $i j k be three coplanar vectors with a b≠ and

v = + +$ $ $i j k, then v is perpendicular to

a. α b. β c. γ d. None of these






one more than one

56. The minimum value of the expression p z z z i= | | + | − | + | − |2 2 23 6 is q, then the value ofq

6is

57. If x , y ∈ R, satisfying the equation( )

,x y−

+ =4

4 91

2 2

then the difference between the largest and

smallest value of the expressionx y2 2

4 9+ is

58. If A B C, , and D are four points in spaces, then AB CD + BC AD + CA BD× × × =k

(area of ∆ABC), where k is equal to

59. If the only integral solution of the equation ( ) ( ) ( ) ( )x x x x− + − + − +…+ − =1 2 3 2007 03 3 3 3 is K

thenK

502

is

60. The equation to the side BC of ∆ABC is x =2. If the altitude through A meets the circumcircle of

∆ABC at P( , )7 2− , then the absolute value of the sum of the x and y-coordinates of the orthocentre

of ∆ABC is equal to





0 9

1. (a) Idea The problem is based on rotating rod in amagnetic field ( )B perpendicular to it, then it’sdifferent parts will be in circular motion withlinear speed v r= ω. So, due to motional emfwe can find out the emf generated between thetwo ends O and A. Then by change in rotation,the polarity of emf will also change.

When the rod is rotating anti-clockwise

e B xdxl

= ∫0ω

=

B

l

ω x 2

02

= Bl

ω2

2[ ]

e V VB l

A1 0

2

2= − =( )

ω

When the rod is rotating clockwise, then

e V VB l

A2 0

2

2= − = − ω

⇒ So, e eB l B l

2 1

2 2

2 2− = − −

ω ω = −B lω 2

TEST Edge In JEE Advanced/IIT JEE, different typesof questions were asked on induced emf. There aredifferent ways by which emf could be induced.Consider one such way as

(i) If the conductor is not in motion but magnetic fieldis varying then low the electrons will feel the forceto induce an emf between the ends of loop.

(ii) We can learn a new thing that a varying magneticfield generates an electric field and due to thiselectric field, the electrons will feel some force andan emf will be induced between the ends.

2. (d) Idea The question is based on circular motion.When sphere is rotating against the wall andfloor, so there will be frictional force on it. Thenormal reaction will adjust itself and due to itthe f Nk k= µ can be adjusted. The sphere willbe in translational equilibrium but it will haveangular retardation.

FBD of arranged object, it does not move it justrotates. So, resultant force will be zero.

N N2 1= µµN N mg2 1+ =

Nmg

1 21=

+ µ

and Nmg

2 21=

+µ

µ

Braking torque

τ µ µ µ µµ

= + = ++

=N r N r mgr mgr1 2 2

1

1

( )( )Q µ =1

τ α= I

2

52mr mgrα =

α = 5

2

g

r(constant)

So, it will come to rest

ω ω α= +0 t

05

20= −ω g

rt

tr

g= 2

50ω

TEST Edge In JEE Advanced, rolling motionquestions may come as it is an important part ofmechanics.

(i) Rolling can be divided into two parts, purerolling and rolling with slipping. Even, we canhave pure accelerated rolling where withv RCM = ω, we have a RCM = α

(ii) In pure rolling, there is no need of any externalforce. But in pure accelerated rolling, externalforces such as friction may also needed.

3. (a) Idea It is based on the concept of moment ofinertia of uniform circular disc and angularacceleration of a disc on a horizontal surface.

In FBD of system,

Retarding angular acceleration fk mg= µ

µ αmgr

mr=2

2, α µ= 2 g

r

For final rest, vf = 0 and ωf = 0


N2

µN2

N1

µN1

mg

v0

ω0

f mgk = µ

v v at v gtf = + = −0 0 µ , tv

g= 0

µ

ω ω α ω µf t

g

rt= + = −0 0

2, t

r

g= ω

µ0

2

So,v

g

r

g0 0

2µω

µ=

v

r0

0

1

2ω=

TEST Edge This concept is important according toJEE Advanced. Every year atleast two questions areasked and students should concentrate on momentof inertia and its rotational mechanism.

Remember if the angular velocity of rotation varieswith the magnitude of tangential velocity as v r= ω.The rate of change of tangential velocity is known astangential acceleration ( )at . Given as

av

tr

tt = = ⋅∆

∆∆∆ω

a rt = α ∴ =

α ω∆∆t

Note ω and α are same for all particles in the bodywhile v and at are different for different points.

4. (c) Idea The problem is based on the combination oftwo long charged wires will act as a capacitorand you have to find the capacitance of the

combination. CQ

V= , where V is potential

difference between two wires. By usingformula it’s capacitance could be found.

Let us give equal and opposite charges to two wiresso that they would have linear charge densities as + λand −λ.

Electric field at point P

Ex a x

= +−

λπε

λπε η2 20 0 ( )

dV Edx Edxa

a x= − = −∫ ∫∫

−η

where, a is radius of wire

⇒ CQ

V= =

| |

π εη0

In

TEST Edge Capacitor is important part ofelectrostatics and question may come fromcapacitors in JEE Advanced. The questions in thecombination of capacitors may also come so let usconsider an example

Here, concentrate on charge distribution and thenapply Kirchhoff’s loop rule to solve the question.

5. (d) Idea We can apply conservation of angularmomentum to solve this problem. Weconsider the angular momentum of the systemwith repect to centre (or any other point) andthen apply conservation of angularmomentum.

See the motion of system (boy + platform) from theaxis of rotation with respect to axis, jump of body willnot develope any external force or torque, so angularmomentum of system will be conserved.

L Li f=I Ii i f fω ω=

(I m I mvR+ = +R 2 )ω ωf

ω ωf

I mR mvR

I= + −( )2

TEST Edge Conservation of angular momentum isimportant for rotational motion and for JEEAdvanced. So, let us consider a situation where wehave to apply conservation of angular momentum.

If a particle hits a box at height h and strikes to itfrom its centre then the angular momentum ofsystem after collision about the point O ′.Note Apply linear momentum conservation firstand then angular momentum conservation.

6. (c) Idea It is based on refraction of light through aglass slab and intensity variation in Young’sdouble slit experiment

i e. . , I I= ′42

2cosδ

where, I is the resultant intensity, I’ is theintensity due to single slit and δ is the angle ofdeviation.

PRACTICE SET 2 87

λ –λP

x

dx

ω

r

Rω

rω

2 Fµ

– q2

4 Fµ

4 Fµ

BA

– q1+q1

2 Fµ

6 Fµ

– q2+q2

+q2

+q1– q1

+( )q – q2 1

– q – q( )2 1

+ –

µ = 0

hM

O'

mv

Y

OX

I IR =

4

20

2cos∆θ

As given IR at centre is 2I0, then

2 42

0 02I I=

cos

∆θ

⇒ ∆θ π2 4

= [for minimum result]

∆θ π=2

∆ ∆x = × =θ λπ

λ2 4

For slab of thickness ∆x t= −( )µ 13

21

4−

=t

λ

t = λ2

TEST Edge The problem is based on refraction oflight, it’s intensity and relate to lateral shiftinterference pattern are important according toJEE Advanced.

Remember In this case, ray MA is parallel to rayBN. But the emergent ray is displayed laterally by adistance d which depends upon µ, t and i and itsvalue is given by

d ti

ii= −

−

12 2

cos

sinsin

µ

where d is lateral shift.

7. (c) Idea It is based on the concept of angular velocityand conservation of angular momentum of abody.

i e. . , L r p= ×where, L is angular momentum.

During the motion of object, there is only oneforce tension which is acting towards a fixed point(centre of table/disc). So that motion is under theeffect of central force and thus, angular momentumwill be conserved.

At any radial distance r is

L m r= ω 2

dL

drmr

d

drm r= + =2 2 0

ω ω ( )

So,d

dr r

ω ω= − 2

TEST Edge This concept is more important accordingto JEE Advanced. Every year one or two questionsare asked and students should relate angularmomentum with torque such as

Angular momentum L r p= ×d

dt

d

dt

d

dt

L rp r

p= × + ×

= × = ×p v r Fm

= + × =0 extr F τ

⇒ d

dt

L = τext

Note Thus, rate of change of angular momentum isequal to the torque due to an external force.

8. (d) Idea This problem is based on the moment ofinertia of a disc and conservation ofmechanical energy

Mechanical energy =Potential energy + All thetypes of energy ( . . ,i e rotational + transitional )

String does not slip relative to disc so onlyconservative force acting in system, so mechanicalenergy is conserved.

loss in PE = gain in KE

mgh mvmr mr= +

+

1

2

1

2 2

1

2 22

22

22ω ω

= + = =1

2 2

2

22

2 2 22mv

mr mvmv

ω

v gh=

TEST Edge These type of problems are important

according to the JEE Advanced point of view. Every

year two or three questions are asked. Students

should focus on kinetic energy of a body in

combined rotational and translation effect.

Remember, a plan motion can be considered ascombination of translational motion of the centre ofmass and rotational motion of body.

Net KE = KT (Translating KE) + KR (Rotating KE)

∴ K mvR

N = +

1

212

2

2

( )KE


d

D

A

i

µ

i

B

N

M

r

t

C

+

ω α,

ω α,

v, a v, a=

Plane motion Translation motion

of centre of mass

Rotational motion

of body about an axis of

rotation passing through

centre of mass

9. (b) Idea By drawing the ray diagram one can easilysolve this question. Here, the focus of lens andcentre of curvature of mirror will coincide. So,the relation between the | |f2 and | |f1 could beeasily found.

The rays incident on the lens will refract as shown andif the focus of lens and centre of curvature of mirrorcoincide then the rays will retrace it’s path and findimage will be at infinity.

So, from the given situation 2 2 1| | | |f f> .

TEST Edge In JEE Advanced/IIT-JEE, the questionsfrom ray optics are very probable. On thecombination of lenses and mirrors differentquestions could be asked.

Let us consider an example If we consider twolenses one is convex and the other is concave thecombination may act as converging or diverginglens depending on their focal lengths. We can find

the effective focal length by1 1 1

1 2F f f= + .

10. (d) Idea This problem is based on rolling friction,torque and angular acceleration.

i e. . ,Torque( )τ = force ×perpendicular distancebetween force and axis of rotation

τ θ= ×r F = rFsin

(where θ is the angle between r and F)

FBD for disc

Equation of force f ma=

τ α− =f RmR 2

2[ α]a R=

τ αR

fm

Rma f− = = =

2 2 2( )

τR

f= 3

2

fR

= 2

3

τ

f N≤ µ(where µ, is coefficient of friction)

2

3

τ µR

mg≤ ( )

µ τ≥ 2

3mgR

TEST Edge Every year, two or three questions arebased on this concept in JEE Advanced. Studentsshould concentrate on work done by torque, powerand angular impulse in rotational motion.

(i) In rotational motion, total work done by a torque isgiven by ω τ θ⋅ ∫D d

(ii) Instantaneous power in rotational motion, p = ⋅τ ω(iii) Angular impulse = = −∫ τdt L Lf i

where Lf and Li are angular momentum of initial

and final position.

In other word, angular impulse of torque is equal tototal change in angular momentum of the body ingiven time.

11. (a,b,c,d) Idea It is based on the conservation ofmomentum we can find the speed of CM.As there is no external force on the systemso, aCM = 0. Finally both the blocks willmove with constant velocity.

vmv m

m

vCM = + =2 0

3

2

3

( )

No external force on system so aCM = 0, vCM willremain constant.

Impulse on m will be in direction of its motion, so it willincrease its momentum.

Impulse on 2m is opposite to the direction of motion,so momentum will be decreased.

TEST Edge CM concept is used in many other topicssuch as rotation, gravitation, so it is important fromJEE Advanced point of view.

In the above case, if we replace string by spring thespring will stretch and some KE of the system willconvert into PE of the system. Even in the case ofstring the string has to be elastic otherwise theabove case will not be possible.

12. (a,c) Idea Just after burning the thread the springforce does not change instantaneously. So justafter burning the thread we can consider theforce and we could easily solve the question.

Spring force does not change its direction andmagnitude instantaneously.

FBD just before the cutting,

Q kx m g= 1

PRACTICE SET 2 89

mga

α

fNτ

f2f1

C1

F1

X = f1

Y = f2 2

k

m g1

m2

T

Tm g2

kxkx

kx

kx

Just before and after cutting acceleration of a will bezero.

ButT becomes zero just after cutting string so motionof m2

kx m g m a− =2 2

m g m g m a1 2 2− =

∴ am m g

m= −( )1 2

2

TEST Edge This problem is the combination ofspring force and constraint motion. The multipleconcept questions are more likely to asked inJEE Advanced. Let us consider one such example.

A ball is projected at speed u at angle θ. At thehighest point it explodes into two equal parts. If onepart reached back to it’s initial projection point Othen the position of second ball where it hits theground can be find out.

This question is based on the combination of CMconcept and projectile motion.

13. (b,c) Idea It is based on the concept of circularmotion. As the box move downwards, itsspeed as well as required centripetal force

mv

R

2

will also increase. Force FC will be

provided by the component of weight mgcosθ.Note carefully that as box is comingdownwards, then FC is increasing and mgcosθis decreasing.

The box will leave the circular surface just after theN = 0 condition, not at N = 0 instant because at N = 0,the component mg cos θ will provide the required

centripetal forcemv

Rmg

mv

R

2 2

⋅ =cos θ at N = 0.

TEST Edge Some of the questions were asked oncircular motion in IIT-JEE and may asked in JEEAdvanced. In most of the circular motion questions,we just have to equate the required centripetalforce and the real force. For example the requiredcentripetal force for moon to revolve is provided byreal gravitational force, so F FC G= .

14. (a,b,c,d) Idea In this problem, first see distribution ofcurrent in the circuit. Use Kirchhoff’s firstrule (junction rule) to see the distributionof current and then use Kirchhoff’s secondrule by considering loops to find the valueof currents and emf of batteries.

After redrawing the circuit

a. lCD = 5 A

b. c. From loop (1),

− + − =8 3 4 3 01( ) ( )E

E1 36= V

From loop (2) 4 5 5 2 8 3 02( ) ( ) ( )+ − + =E

E 2 54= V

d. From loop (3), − − + =2 01 2R E E

RE E= − = − =2 1

2

54 36

29 Ω

TEST Edge The Combination of resistances is animportant topic of current electricity and it maycome in JEE Advanced from this topic.

There are two to three methods that we use to solve thistype of problem.

1. By Wheatstone bridge method

2. By Kirchhoff’s rule

3. By removing the parts of circuit if the potentialdifference between the ends of that part is zero.In this case, there will be no current in that partof circuit.

15. (b,d) Idea The problem is based on work-energytheorem.

Work done by all the force acting on a body isequal to change in kinetic energy of the samebody.

i e. . , ∆KE = netW


2 A

4 Ω3 Ω 2 Ω

4 Ω

3 A

E1 E2

R

1 A

(1) (2)

5 A

7 A

10 V

+

–

(3)

5 A + 3 A 5 A

DC

+ – – +

2 A

4 Ω3 Ω 3 Ω 6 Ω

4 Ω

3 A

E1 E2A

10 V

+ +

BC D

R

m/2

m/2

m

O

θu

FC

θ

mg acosmg sin θ

mg

N

θv

Loss in the energy of photon = − −h h dν ν ν( )

= h d( )ν= work done by gravity

Let say m = mass of photon = h

c

ν2

hd duν = −u

GMm

xdu

dx

GMm

x

= −

= +

2

hdGMm

xdx

GMh

c

xdxν

ν

= − = −

2

2

2

hdGMh

c xdxν ν= −

2 2

d Gm

c

dx

xR

R Hννν

ν

0

= −∫ ∫+

2 2

hGM

c R h R

GMh

R c

νν0

2 2 2

1 1

= +

+−

= −

ν ν=−

0

2 2

e

GMh

R c

ν ν=−

02

e

gh

c QgGM

R=

2

TEST Edge It is important according to JEEAdvanced. Every year one question is asked andstudent should focus on work energy theoremand relate to power and potential energy of aspring.

Remember This theorem can be applied tonon-inertial frame also in an non-inertial frame itcan be written as work done by all the force(including the pseudo forces) change in kineticenergy in non-inertial frame.

The slope of work-time curve gives the

instantaneous power as pd

dt= =ω θtan as shown in

figure.

16. (3) Idea The electron of hydrogen atom will gainsome part of KE of the system and it will jumpto the higher state. While returning to theground state it will emits photons. Byobserving the excited state we an find thepossible number of photons that could beemitted.

Maximum possible loss of energy isKE

2in , when

collision is perfectly inelastic as shown.

∴ (KE )loss max = =25 5

21275

.. eV

This is sufficient to raise the electron upto 3rd shell.While returning to first shell three photons could bepossibly emitted.

TEST Edge Atom is the part of modern physics andthe questions from modern physics are relativelyeasier than mechanics or electricity and magnetism.So, it is easier to score marks from this part. So,concentrate on this part as well and remember itsformulae and theory to solve its questions quickly.

17. (3) Idea Here don‘t be confused with the choice ofzero potential. As W Vext = ∆ does not dependon the choice of zero potential. Just apply theformula for Wext with considering zeropotential at infinity and you will get theanswer.

When we consider the value of potential to be zero at

infinity then formula of potential−GM

re

⇒ W A B V V mB Aext ( ) [ ]→ = −

= − −

GM

R

GM

Rme

e

e

e2

= −

3

2

GM

Rme

e

by comparing it−

x GM m

Re

e2

Don’t be confused by the choice of zero potentialbecause we want to find Wext which depends on ∆V

and ∆V in not depend in the choice of zero potential.

TEST Edge From gravitation, the questions based onconservation of mechanical energy may asked inJEE Advanced. Let us consider one example.

PRACTICE SET 2 91

m

+

Neutron

m

H-atom

⇒2m

u/2

θ

Wo

rk

Time

A

B

Re

mm

r

(i) Two masses are at a distance r as shown in the

figure. Initially they were at rest. Now if they attract

each other with gravitational force then their speeds

when they are at a distancer

2with each other can be

find out.

(ii) Just apply conservation of mechanical energy to

solve it.

18. (0) Idea The problem is based on impulse andconservation of energy in elastic collision.

Ball with respect to surface

Upward impulse on ball

Ndt p pf i= −∫Ndt v y= − − ×∫ 2 2 4( )

Ndt v y= +∫ 2 8 ...(i)

Equation of e along normal direction

ev y=4

...(ii)

v ey = =4 2 m/s

So, Ndt = + =∫ 4 8 12

Horizontal impulse

− = −∫ µNdt p pf i

– µ Ndt v x= − ×∫ 2 2 2

− × = −1

312 2 4v x

v x = 0

TEST Edge It is important according to JEEAdvanced. Every year one question is asked andstudent should concentrate in elastic, inelasticcollision and impulsive force.

Remember In a perfectly elastic collision,

velocity of separation = velocity of approach andfor a perfectly inelastic collision,

velocity of separation = 0

(i) If coefficient of restitution (e) is equal to 1, thencollision is perfectly elastic and if e = 0, thencollision is perfectly inelastic.

(ii) The change in momentum produced by such animpulsive force is

p p pp

Ff ip

p

t

t

t

t

i

f

i

f

i

fd

d

dtdt dt− = = = ⋅∫ ∫ ∫

19. (3) Idea The problem is based on energy stored in acapacitor and to calculate amount of heatenergy stored when capacitor is filled with adielectric constant.

Initial charge on capacitor = CE

Initial potential energy of capacitor = CE 2 2/

Now, C A d= ε0 /

and new capacitance CA

d′ = ε0

2/, in series with

ε0KA

d /2

⇒ =CA

d′ 2 0ε

series4 0ε A

d

= ×+

= =ε ε0 02 4

2 4

4

3

4

3

A

d

A

dC

⇒ New charge = 4

3CE

New energy = ×1

2

4

32CE

= 2

32CE

Now, W battery = +∆ ∆H U

⇒ E CE CE H CE CE4

3

2

3

1

22 2−

= + −∆

⇒ ∆H CE= = × × =1

6

1

62 3 32 2( ) J

TEST Edge This concept is generally asked in JEEAdvanced. Student should relate to energy densityin a electric field and parallel plate capacitor with adielectric.

Case 1. If a conducting slab ( )k = ∞ is placed between

the plates, then

CA

d tt

A

d t=

− +∞

=−

ε ε0 0


2 m/s

(2 + 2) = 4 m/s2

vx

vy

N

µN

Time

Impulsive force

Fo

rce

ti tf

K

Case 2. If the space between plates is completely filledwith a conductor, then t d= and k = ∞

then, CA

d dd

=− +

∞

= ∞ε0

where C is capacitance.

20. (2) Idea It is based on the concept of velocity oftransverse wave and centrifugal force on astring.

Velocity of transverse wave on rope is = T

µwhere,T = tension in string.Variation in tension in a horizontal rotating string.

For small element of rope

T T dT dm r− + =( ) ω2

− =dT dm xω2

− = ∫∫ dT xdxx

T

Tµω2

00

T Tx− = −0

2 2

2

µ ω

T Tx= −0

2 2

2

µ ω

At x L T= =, 0,

02

0

2 2

= −TLµ ω

TL

0

2 2

2= µω

TL x= −µω2 2 2

2

( )

So, velocity of wave on string

vT L x L x= = − = −µ

µωµ

ω2 2 2 2 2

2 2

( )

dx

dtL x= −ω

2

2 2

dx

L xdt

L t

2 20 0 2−=∫ ∫

ω

sin−

=1

0 2

x

Lt

L ω

sin ( )− = =1 12 2

π ωt

∴ t = πω2

As given in questionπ

ωπ

ω2=

n⇒ So, n = 2

TEST Edge It is important according toJEE Advanced, every year one question is askedfrom this concept students should focus intransverse of waves.

Remember While wave travelling though thestring, then tension (T) is developed in a string isgiven by

T v= 2 2µ

⇒ vT=µ

, where v is speed of wave

when string is whirled, there is a centrifugal forceon the string as whole, which is given by

∴ Fcentrifugal = m rω2

where, r = radius of a circle.

ω =angular speed at each and every section of thebody of the string.

21. (b) Idea Problem is based on concept ofdetermination of age of ore. While solving theproblem students are advised to go throughfollowing steps.

Calculate the total amount of U238 decay.

Then calculate the total amount of U238

present and then after age of ore usingfollowing equation

tN

N= 2 303 0.

logλ

Considering that the whole ofPb206 comes fromU238,

Amount of U238 decayed = 01 206. g Pb

= ×01 238

206238.

gU

= 01155 238. g U

Hence, initial amount of U g U238 2381 01155= +( . )Value of disintegration constant can be calculated as

Now, λ = =×

0 693 0 693

0 5 101 29

. .

./T yrs

= × − −0154 10 9. yrs 1

PRACTICE SET 2 93

ω

dx dm,x

T T dT+

m

Centrifugal force

on body.

TO

mg

v

Conductor

tN

N= 2 303 0.

logλ

=×

−

2 303

0154 10

11155

19

.

.log

.

= ×7 099 108. yrs

TEST Edge Problem based on concept of half-lifetime and isotopic composition can also be asked inJEE Advanced while solving this type of problemwith conceptual mixing of more than two terms,students must calculate important parameters firstfollowed by other problems such as

A mixture containing 239 Pu Puand 240 has specificgravity of 6 0 109. × dis/s. The half-lives of the isotopes are2 44 104. × and 6 58 103. × years respectively then whatwill be the isotopic composition of these samples.

Answer = =. , :239 240 39 61Pu Pu

This problem can be solved by calculatingconcentration of both isotopes separately and thentheir ratio by using concept of mole fraction.

22. (c) Idea Problem is based on concept of Einsteinequation of photoelectric effect while solvingthe problem students are advised to use theEinstein equation.

According to the Einstein equation of photoelectriceffect

h hν ν= +0 KE

∴ 1

22

0mv h h= −ν ν

= −hc hc

λ λ 0

vhc

m

2

0

2 1 1= −

λ λ

vhc

m= −

2 0

0

λ λλλ

Hence, (c) is the correct choice.

TEST Edge Similar type of problem based onconceptual mixing of photoelectric effect,de-Broglie equation and Hisenberg uncertainityprinciple can also be asked in JEE Advanced.

23. (b) Idea This problem includes concept ofpolymerisation. While solving the problemstudents are advised to undergo knowledge ofcarbocation as well.

More will be the stability of carbocation produced bymonomer, more will be the extent of cationicpolymerisation.

Since, cationic polymerisation proceeds throughformation of carbocation hence most stable monomerfor cationic polymerisation is that which producesmore stable carbocation.

Intermediate of cationic polymerisation will becarbocation.

(a) CH CH C

O

OMe3

+

(b) CH C

H

O C

O

CH3 3

+

most stable carbocation

(c) CH CH Cl3

+

(d) CH CH CN3

+

Hence, correct choice is (b).

In option (b)

O C

O

CH3 group stabilise

carbocation.

TEST Edge Problems based on properties and usesof important polymers such as nylon, Teflon, PVCcan be asked in JEE Advanced.

24. (d) Idea This problem is based on conceptual mixingof isomerism and molar conduction ofcoordination compound. While solving theproblems it is advisable to go through theunderstanding of isomerism and vant Hofffactor of coordination compound.

Out of given options, choices (d) contain[CoBr (NH ) ]SO2 3 4 4 and [Co(SO )(NH ) ]Br4 3 4 2 both areionization isomers and when they dissolve in aqueoussolution produces different number of ions inaqueous solution as

[CoBr (NH ) ]SO [CoBr (NH ) ] SO2 3 4 4 2 3 42

42

s+ −+

Total charges = 4

[Co(SO )(NH ) Br ]Br [Co(SO )(NH ) ]4 3 4 2 2 4 3 4s+

+ −2Br

Total charges = 3

Since, charges are different in a aqueous solutionhence, they have different molar conductivities.

TEST Edge Similar problems based on conceptualmixing of isomerism and qualitative analysis ofinorganic compounds can also be asked in JEEAdvanced the such as.

What is the magnetic moment of complex used inqualitative analysis of nitrate ion?

After determining magnetic moment using theformula

µ = +n n( )2 BM.

One can get answer as 1 73. BM.

25. (a) Idea This problem is based on conceptualefficiency of Carnot engine and diagramrepresentation of Carnot engine, while solvingthis problem students must have a goodemphasis on analysis of diagram whichrepresents Carnot engine.


η Area bounded by the curve

Total area under line=

BC×100

η = ×1

2area of triangle× =100 33 33. %

TEST Edge Similar problems based on conceptualbasis of Carnot engine and their differentrepresentation alongwith determination of efficiencyof engine can also be asked in JEE Advanced, sostudents are advised to go through in depth study ofCarnot engine and their representation.

26. (a) Idea This problem is based on concept ofpreparation of sulphur in accordance withformula of Engle’s sulphur. Students areadvised to solve this problem by using correctmethod for preparation of Engle’s sulphur.

Molecular formula of Engle’s sulphur is S6 (alsoknown as cyclohexasulphur). It can be prepared byreaction of S Cl2 2 with H S2 4.

S Cl H S S + 2HCl2 2 2 6+ →4

Structure of S6

While other reaction produces different allotropicforms S as follows

2HNO H S S + 2HNO H O3 2Colloidalsulphur

3 2+ → + 2

2H S SO O + 3S2Colloidalsulphur

+ →2 22H

TEST Edge Problem based on conceptual mixing ofpreparation, properties and structure of differentallotropes of phosphorous can also be asked. So,students are advised to go through in depth studyof these topics.

27. (b) Idea This problem is based on concept ofreaction of phosphorus and their compoundsand their chemical properties.

The sequence of given reaction is as follows

Here, A is PH3 (hydride of phosphorus)

B is P2O5 (pentaoxide of phosphorus)

TEST Edge Problems based on compounds of siliconalongwith their structurs can also be asked hencestudents are advised to go through the study ofsilicon and silicates including their chemicalpreparation.

28. (d) Idea This problem is based on the concept ofintramolecular nucleophilic addition reaction.While solving this problem, students areadvised to use the reaction of carbocation withinternal nucleophile.

TEST Edge Similar problems based onintramolecular rearrangement reaction involvingvarious types of reaction intermediates such asnature of carbon etc., can also be asked inJEE Advanced, so students are advised to gothrough the study of these topics.

29. (a) Idea This problem is based on conceptual mixingof determination of emf of cell using values ofKsp and standard reduction potential.

As we know from Nernst’s equation

E En

cell sp° °= − 0 0592

10.

gK

V E= −° 0 0592

210

.g Ksp

E V° = + 0 0592

210

.gK (QKsp = K)

Hence, (a) is the correct choice.

V K+ 0 059

2

.log

TEST Edge Similar problem based on Nernst’sequation and its application can also be asked sostudents are advised to go through the topic.Problem such as.

Find the emf of the cell for P H H Ptt, H H2 2| | | ,∆ +

1 atm (100 mL of 0.1 m) (0.1 m) (1 atm)

Given that emf of cell = 0 118. V

T = °25 C

if 30 mL of 0.2 M NaOH is added to the negativeterminal of battery.

After solving one can get answer

E° =cell 0 246. V

30. (d) Idea This problem is based on concept ofoxidation reaction of alkenes. Students areadvised to go through the basics of oxidationreaction.

PRACTICE SET 2 95

H⊕

OH H—O ⊕

H—O⊕

–H⊕

O

P3NaOH + 3H O2

PH3

HNO3P O2 5

Oxygen, DA B

S

S S

S S

S

TEST Edge Similar problem based on conceptualmixing type of reaction, intermediate involved inthe reaction and rearrangement of intermediate canalso be asked so students are advised to go throughclear understanding of these topics and relatedproblems such as

What will be the intermediate involved in Favorskirearrangement after undergoing through propermechanism?

One can get carbanion as answer.

Remember It also includes formation ofcyclopropane ring inside as intermediate duringproduct formation.

31. (a,b) Idea This problem includes conceptual mixingof hybridisation and percentage s characterwhile solving the problem, students areadvised to calculate the hybridisation firstfollowed by percentage s character inmolecule, % s character and hybridisation areinter related as

Hybridisation % s Character

sp 50%

sp2 33%

sp3 25%

sp d3 20%

sp d3 2 17%

Molecules Hybridisation s character

I. CO 32− sp2 33%

II. XeF4 sp d3 2 17%

III. I3− sp d3 20%

IV. NCl3 sp3 25%

V. BeCl2 sp 50%

Hence, correct order is II < III < IV < I < V

TEST Edge Similar problems including conceptualmixing of hybridisation, percentage s character,percentage pcharacter, bond angle of molecules canalso be asked in JEE Advanced, such as

What will be the hybridisation of non-bonding pair ofelectrons in NH3 if bond angle is 107°.After solving, you will get answer as sp2 13. .

32. (a,c) Idea This problem is based on structure of

oxides of nitrogen. Students are suggested todraw the structure of molecules before solvingsuch questions be careful during drawing thestructure the oxidation state and number ofelectrons distributed in compound and checkthat these must be in proper ratios.

Structure of oxides of nitrogen

TEST Edge Similar problems based on structure ofoxo acids of sulphur can also be asked in JEEAdvanced, such as

Which of the following doesn’t have s-s linkage?

(a) S O2 82- (b) S O2 6

2− (c) S O2 52− (d) S O2 3

2−

After drawing structure, one can get a as answer.

33. (a,b,c,d) Idea While solving this problem, studentsmust have knowledge about conceptinvolved in rate of esterification, rate ofdehydration, so the reactivity ofderivatives of carboxylic acid, students aresuggested to go through the study ofinductive effect, mesomeric effect,resonance effect, hyperconjugation etc.,which will help them in the solving ofthese types of problems.

(i) Rate of esterification with respect to alcohol∝nucleophilicity of alcohol.

(ii) Rate of dehydration ∝ stability of carbocation,

(iii) Rate of esterification with respect to acid ∝electrophilicity of R

C

O

group.

(iv) Reactivity of acid derivatives

∝ electrophilicity of R

C

O

group.

TEST Edge Problems based on chemical reactivity oforganic compounds such as carboxylic acid andtheir derivatives towards various reducing agentscan also be asked in JEE Advanced. So, students aresuggested to go through in depth study of thesetopics, such as


N—NO

O

O

O

Structure of N O

1 N—N bond2 4

NO

O

O

O

Structure of N O

No N—N bond2 5

ON

N—NO O

Structure of N O2 3

ON==O

O

Structure of NO2No N—N bond1 N—N bond

CH —CH==CH3 2

Br2

CCl4CH —CH—CH3 2

BrBr

(oxidation

reaction)

OH

KMnO4

CH —CH—CH3 2

OHOH

(oxidation

reaction)

H /H OÅ2 CH —CH—CH3 2

HOH

(Addition of both

hydrogen and oxygen)

–

What will be the relation between products of givenchemical reactions?

Knowing the mechanism of Lindlars catalyst Birchreduction, one can get answer as geometricalisomers.

34. (a, d) Idea This problem is based on the concept ofanalysis of NH4

+ and K+ . To solve thisproblem, students must have theknowledge of group reagents used inqualitative analysis of ammonium ion.

Analysis of NH4+ NH4

+ may be analysed by reactionof solution containing NH4

+ with sodiumhexanitrocobaltate (III), which results as yellowprecipitate.

NH4+ + →Na Co NO3[ ( ) ]2 6 Yellow ppt.

Solution containing NH4+ on reaction with H [PtCl ]2 6

produces yellow precipitate also.

NH H (PtCl )4 2 6+ + → Yellow ppt.

Similarly, K+ also produces yellow precipitate onreaction with Na [Co(NO ) ]3 2 6 and H [PtCl ]2 6

K Na [Co(NO ) ] K [Co(NO ) ]+3 2 6 3 2 6+ → ↓

Yellow ppt.

K H [Pt Cl ]2 6+ + → Yellow ppt.

While Na+ and Mg2+ does not give this test.

TEST Edge Similar problems based on concept ofgroup reagents and their qualitative analysis canalso be asked.

35. (a,b) Idea This problem includes the concept ofcharacteristics and nature of amino acidunder different conditions. While solvingthis problem students are advised to studythe statements very carefully for choosingthe incorrect statements. Hence, tick thetrue and false statements first according toinformation provided, then answer thequestion as shown below

As alanine is a neutral non-essential amino acid.Hence, answer of this question can be done as markThe statement as

For true statement ()

For false statement (×)

According to the questions

(a) It is acidic amino acid (×)

(b) It is an essential amino acid (×)

(c) It will move to cathode in electrolysis at pH = 5 ()

(d) It is an optically active amino acid ()

Hence, incorrect statements are (a) and (b) thenanswer is (a) and (b).

TEST Edge Problems based on characteristics ofamino acids as well as preparation and uses ofvarious amino acid can also be asked in JEEAdvanced, related such as

Which of the following processes can be used duringpreparation of amino acid?

(a) Gabriel pthalimide process(b) Strecker synthesis(c) Sandymeyer process(d) All of the above

It can also be asked, students can answer suchquestion if he has the knowledge of amino acidprecursur and proper mechanisms of reaction.

The answers are (a) and (b).

36. (1) Idea This problem is based on concept of δ -bondformation which can be solved by using thebonding involved in δ -bond formation.

For formation of δ -bond, the lobes of orbital must bein phase i e. ., in overlapping condition to each otheras shown below

37. (9) Idea This problem includes the concept ofdetermination of pH using value of Ksp. It isadvisable to go through the study ofequilibrium constant, solubility product andpH determination to solve all these types ofproblems.

Mg(OH) Mg OH22+ + −2

1 10 12× =− + −[Mg ][OH ]2 2

(OH ) 10 M5− −=pOH = − =−log10 55

pH = − =14 5 9 [ ]QpH pOH+ =14

pH = 9

PRACTICE SET 2 97

A

BNa/NH3

H2

Pd/BaSO4

y-axis

x-axis

z-axis

is overlapping a not possible overlapping possible

overlapping possible

undergo bond formation

dxy

δ

δ-bond

z-axis

zy

x

y

x

TEST Edge Similar problems based on conceptualmixing of pH, Ksp electrode potential can also beasked in JEE Advanced, students are advised to gothrough the basic concepts of these topics and relateall these term.. In order to approach towards thecorrect answer, students must follow stepwiseapproach.

38. (6) Idea This problem is based on the concept ofstructure of octahedral compound andnumber of possible type of XMX bond angles,while solving problems, students are advisedto consider all the X as different ligands.

All 12 bond angles are shown by numeric below.

∴ x =12

x

2

12

26= =

TEST Edge Similar structure and hybridisationrelated problems such as

What is the ratio of σ and π-bonds in B N Cl3 3 3?

It can also be asked students, can answer thequestion by drawing the structure of inorganiccompound and counting number of σ and π-bondsand finally calculating the exact ratio, one can getanswer as 3.

39. (8) Idea This problem can be solved by using theconcept of entropy of mixing. Since, twogases are mixing, so students are advisedto use the following formula to solve theproblem.

∆ ΣS R n xi isys = − ln

where, ni = number of moles of gas

xi = mole fraction of gas.

∆ ΣS R n xi isys = − ( ln )

= − × + ×

R 11

32

2

3ln ln

= +

R ln ln3 23

2

= × + ×8 314 1 2 0 4. ( . ) =15 9. J/K ~− =16

28

TEST Edge Problems based on energy of mixing oftwo solids two immiscible liquids, one liquid andone gas can also be asked. So, students are advisedto go through the concepts of these topics.

40. (5) Idea This problem includes the concept ofLindmann theory of unimolecular reactionswhich is used to calculate apparent rateconstant. If is advisabli to use the conceptof limit and continuity (mathematics) tosolve this problem.

90

100 1 11 1×

+

=

+→∞lim

c

k c

c

k c

cα α

90

100 1 11 1×

+

=+→∞

limc

k

c

k c

cα α

⇒ 90

100 11 1× =

+k k c

cα α⇒ 0 9

1

.

α α=

+c

c

0 91

. =+α

αc

c

⇒ 0 9 0 9. .+ =α αc c ⇒ 01 0 9. .αc = ⇒ α c = 9

c = =×

9 9

9 105α⇒ c = −10 5 M. ∴ | log |c = 5

TEST Edge Similar types of problems, withconceptual mixing of mathematical concept andrate constant determination can also be asked in JEEAdvanced. Some problems including concept ofClausius theory and transition state theory can alsobe asked.

41. (a) Idea If ax bx c2 0+ + = is a quadratic equation,

then sum of roots ( )α β+ = − b

aand product of

roots ( )αβ = c

aand arithmetic mean A of any

two numbers a and b isa b+

2.

Given equation is x x2 1 0− − =⇒ α β+ = 1, αβ = −1

Now, Ann n= +α β

∴ Ann n

−− −= +11 1α β

Hence, AM =+

= + + +−− −A An n

n n n n1

1 1

2 2

α β α β…(i)

Now, Ann n

++ ++1

1 1= α β= + + − +− −( )( ) ( )α β α β αβ α βn n n n1 1

Ann n n n

+− −= + + +11 1α β α β …(ii)

[ , ]Qα β αβ+ = = −1 1

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

AM = +An 1

2

TEST Edge Questions based on quadratic formula tofind the roots of the equation and concept related toprogression (AP, GP and HP) are asked. So, to solvesuch questions, students are advised to learn natureof roots, relation between roots and coefficient ofquadratic equation and basic facts of arithmeticprogression such as nth term of AP i e. . ,a a n dn = + −( )1 , sum of nth term of AP i e. . ,

Sn

a n dn = + −2

2 1[ ( ) ] etc.


M

X

X

8

65

2

10

11 9

12

4

3 1

7

X

X

XX

42. (c) Idea We have three inverse functions

sin ( )− −1 1x , cos ( )− −1 3x and tan−

−

1

22

x

x

whose domain will be [–1, 1], [–1, 1] and( , )− ∞ ∞ respectively and from the value of

trigonometric function cosπ4

1

2= . We find the

value of k.

For sin ( )− −1 1x to exist x − ∈ −1 1 1[ , ]

x ∈ [ , ]0 2 ...(i)

For cos ( )–1 3x − to exist

x − ∈ −3 1 1[ , ]

λ ∈ [ , ]2 4 ...(ii)

Eqs. (i) and (ii) ⇒ x = 2

So, sin ( ) cos ( ) tan− − −− + − +−

1 1 12 1 2 32

2 4

= +−cos 1k πsin ( ) cos ( ) tan ( ) cos− − − −+ − + − = +1 1 1 11 1 1 k π

π π π π2 4

1+ − = +−cos k

π4

1= −cos k ⇒ k = cosπ4

k = 1

2

TEST Edge Domain and range of inverse functionand value of trigonometric functions basedquestions are asked. So, to solve such types ofquestions students are advised to learn domain andrange of various trigonometric functions such astan− 1 x, cosec−1 x, sec− 1 x, etc., and its graphicalrepresentation and also acquainted yourself withthe values of trigonometric functions at the given

angle e.g., tan cot− −=

1 1 1x

xif x > 0 and

tan cot− −= − +

1 1 1x

xπ if x < 0 because range of

these two functions are different.

43. (a) Idea If a i j k= + +a a a1 2 3$ $ $ and b i j k= + +b b b1 2 3

$ $ $

then, a b⋅ = + +a b a b a b1 1 2 2 3 3 ,

| |a = + +a a a12

22

32 and | |b = + +b b b1

222

32

and any point which lie on the curve satisfiesthe curve equation.

Let OA i+ j= x y1 1$ $ and OB i j= +x y2 2

$ $

Since, OA i⋅ =$ 2

x1 2=Also, OB i⋅ = −$ 3

⇒ x2 3= −Since, the points (x y1 1, ) and ( , )x y2 2 lie on curve.

∴ y x1 12= and y x2 2

2=When x1 2= , y1 4=When x2 3= − , y2 9=∴ OA i j= +2 4$ $

OB i j= − +3 9$ $

Now, 3 4 3 2 4( ) ( ) ( $ $)OA OB i j− = + − −4 3 9( $ $)i+ j

= −18 24$ $i j

∴ | | ( ) ( )3 4 18 242 2OA OB− = +

= +324 576

= =900 30 units

Required length is 30 units.

TEST Edge Questions based on basic coordinategeometry, position vector and linearly dependentvectors are asked in conceptual mixing. So, studentsare advised to understand basics of coordinategeometry and also acquainted yourself withconcept of vectors such as dot and cross products oftwo vectors, types of vectors like position vector,coplanar vector, etc. e g. . , if a line passes through thepoint A( )a and is parallel to the vector b, then its

equation is r a b= + t .

44. (a) Idea Modulus function y f x x= =( ) | | is defined

as yx x

x x=

≥− <

, if

if

0

0,

Range of the trigonometric function cos x is Rand properties of definite interval

a

b

a

c

c

bf x dx f x dx f x dx∫ ∫ ∫= +( ) ( ) ( ) . Also,

standard result of cos sinx dx x C∫ = + and

general solution of trigonometrical equation ifsin sinθ α= , then θ π= + −n n( )1 α where,

α π π∈ −

2 2

, .

If 02

< <xπ

, | cos | | cos |/

x dx x dxx

+− −

−

∫ ∫2 2

2π

+−∫ | cos |

/x dx

x

π 2

⇒ cos cos/

– /x dx x dx

x

−

−

∫ ∫+2

2

2

π

π+ | cos |x dx

x

=−∫ 02

⇒ | sin | | sin |/

− + − =−

−

−x x x

2

2

2

0π

π

⇒ 1 2 1 0− + + =sin sin x

⇒ 2 − + =sin sin2 0x

⇒ sin sinx = − < −2 2 1 not possible

∴ No solution exist in 02

,π

So, number of solution = 0.

TEST Edge In JEE Advanced, questions based ondifferent types of functions such as trigonometricfunction, modulus function, greatest integerfunction and concept related to properties ofdefinite integral and general solution oftrigonometrical equations are asked. So, to solvesuch types of questions learn definition of variousfunctions, different properties of definite integraland methods to find general solutions oftrigonometrical equation e g. . , general solution oftan tan2 2θ α= is θ π α= ±n , n I∈ .

PRACTICE SET 2 99

45. (a) Idea If f x f x( ) ( )− = ∀ x in the domain of f then f

is an even function and if f x f x( ) ( )− = − thenf is an odd function. If f A B: → , then domain

of f a a A a f a f= ∈ ∈ / , ( , ( , )) and range off f a a A f a B= ∈ ∈ ( )| , ( )

x f x fx

g x2 21

( ) ( )−

= …(i)

Replacing x by1

x, we have

1 12

12x

fx

f x gx

− =

( )

Multiplying by 2 2x

21

4 212 2f

xx f x x g

x

− =

( ) …(ii)

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

− × = +

3 2

12 2x f g x x gx

( ) ( )

⇒ f x

g x x gx

x( )

( )

= −+

21

3

2

2

Now, f xg x x g x

x( )

( ) ( / )− = − − + −

2 1

3

2

2

= +

g x x g x

x

( ) ( / )2 1

3

2

2

∴ f x f x( ) ( )= − −f x( ) is an odd function.

But f x( ) is given to be an even function.

∴ f x x f( ) ( )= ∀ ⇒ =0 5 0

TEST Edge Questions based on domain and range ofdifferent types of functions and their value in anyinterval are asked in JEE Advanced. So, to solvesuch types of questions, students are advised tolearn definition of different functions with rangeand domain such as trigonometric functions, evenor odd function, etc. e.g., iff x b x bx( ) ( )= + + +1 2 12 2 and m b( ) is the

minimum value of f x( ) then for the range of m b( ),we need to find minimum value and the range ofgiven function f x( ), so range of m b( ) is (0, 1].

46. (d) Idea For any complex number z, we have Re (z)as − ≤ ≤| | Re ( ) | |z z z and Im( )z as− ≤ ≤| | Im ( ) | |z z z and for any function f x( ) iff ’( )a = 0 and ′′ <f a( ) 0, then f x( ) would have a

local maximum at x a= , otherwise when

′ =f a( ) 0 and ′ ′ >f a( ) 0, then f x( ) would havea local minimum at x a= .

Let z a ib= + , a > 0, b > 0

Since, Re( ) Im( )z z+ = 3

∴ a b+ = 3

Let P z z= (Re ) Im ( )2 = a b2

P a a= −2 3( )

= −3 2 3a a (to be maximised)dP

daa a= −6 3 2

For maxima or minimadP

da= 0

⇒ 6 3 02a a− =⇒ 3 2 0a a( )− =⇒ a = 0 or a = 2

Now,d P

daa

2

26= −

d P

daa

2

22

12 0

= − <

=

∴ a = 2 is point of maxima.

Maximum value ofP = −( ) ( )2 3 22 = 4

TEST Edge Questions based on the expression ofcomplex number and related to maxima or minimaof function, students are advised to learn the basicrepresentation of complex numbers in terms of realand imaginary numbers also acquainted yourselfwith second order derivative test to find themaximum or minimum value of the function. e.g.,the minimum value of functionf x x aa x( ) log log= + is to be verified by the secondorder derivative test so you will find, it is false.

47. (d) Idea Chain rule of differentiation if y f g x= ( ( ))

and u g x= ( ), thendy

dx

dy

du

du

dx= ⋅ and the concept

of definite integral0

bf x dx F b F a∫ = −( ) ( ) ( ),

where f x dx F x C( ) ( )= +∫

Considerdf x

dx

e

x

x( ) cos

=

⇒ df x

dx

e

x

x( ) cos3

3 3

3

=

⇒ df x

dx

d x

dx

e

xx

x( ) ( ) cos3

3

3

32

3

3= ⋅

⇒ df x

dx

e

x

x( ) cos3 33

= ...(i)

Now, consider the given integral

31

3

1

5e

xdx f k f

xcos

( ) ( )∫ = −

⇒ d f x

dxdx f k f

[ ( )]( ) ( )

3

1

5

1= −∫ [using Eq. (i)]

⇒ d f x f k f[ ( )] ( ) ( )3

1

5

1= −∫⇒ [ ( )] ( ) ( )f x f k f3

15 1= −

⇒ f f f k f( ) ( ) ( ) ( )125 1 1− = −⇒ k =125


TEST Edge In JEE Advanced, questions based ondifferentiation rule and basic concepts related todefinite integral are asked. To solve these types ofquestions learn the basic formulae of differentiationsuch as quotient rule, product rule, etc., and basicdefinition of definite integral as area function etc.

eg., if0

1x

xf t dt x t f t dt∫ ∫= +( ) ( ) , then the value of

f ( )1 is obtained by differentiating w.r.t. x, so

f ( )11

2= .

48. (c) Idea If y ax bx c= + +2 is an quadratic equation,

then value of discriminant, D b ac= −2 4 and

the graph between x and y is always aparabola. If a > 0 and D < 0, then shape of theparabola is concave upwards ∀ ∈x R, y > 0.Probability of an event is given by

P E( ) = Favourable outcomes

Total outcomes

Since, the curve is strictly above the x-axis

∴ x a x a2 2 4 5 64 0+ + − + ≥( )

⇒ D < 0

4 4 4 5 64 02( ) ( )a a+ − − + <⇒ a a2 13 48 0+ − <⇒ ( )a + 16 ( )a − 3 <0

⇒ − < <16 3a

Also, a ∈ −[ , ]5 30

∴ a ∈ −[ , ]5 2

∴ Number of favourable cases = 8

Total number of values of a in [ , ]−5 30 i.e., 36

∴ Required probability = 8

36= 2

9

TEST Edge Questions based on quadratic equationsuch as roots of the equation, discriminant value,nature of roots and basic definition of probabilityare asked. So, to solve such types of questionsunderstand the results of quadratic in equalities andits graphical representation in different shapes ofthe parabola and also understand the application ofthe probability definition.

For any quadratic equation ax bx c2 + + , if D = 0then roots are equal i e. . , x x1 2= in that case a > 0

⇒ x x x∈ −∞ ∪ ∞( , ) ( , )1 1 and a < 0

⇒ x ∈φ

49. (b) Idea If a line r a b= + ′λ lies in the plane r n⋅ = d ,

then b n⋅ = 0 and a n⋅ = d, vector equation of astraight line passing through a position vectora and parallel to a vector b is r a b= + λ .

Equation of line passing through

a and ⊥ to r n⋅ = d is r a n= + λ

For foot of perpendicular,

( )a n n+ ⋅ =λ d

⇒ a n n⋅ + =λ | | 2 d

⇒ λ = − ⋅d a n

n| |2

∴ Foot of perpendicular is aa n

nn+ − ⋅

d

| |2

∴ Equation of line parallel to r a b= + λ and passingthrought foot of perpendicular

r aa n

n= + − ⋅

d

| |2n b+ λ

TEST Edge Questions based on planes, lines and alsoconcept related to lines with planes such as anglebetween them, distance between them, vector form,Cartesian form of lines and planes are asked. Tosolve such types of questions, students are advisedto understand the concept of lines and planes suchas length of the perpendicular from a point havingposition vector a to the plane r n⋅ = d is given by

Pd

=⋅ −a n

n| |need to use and understand.

50. (b) Idea Let y f x x= =( ) [ ] is a greatest integer

function. If n x n≤ < + 1, then [x] = n, i e. . , [2.3] =2, [4.3] = 4. Now, apply the concept of maximaand minima, then take lim x → 0, to solve thisproblem.

y y y2 4 11 2 7− + = − +2( )

Now, ( )y − ≥2 02

( )y − + ≥2 7 72

∴ Minimum value of y y2 4 11 7− + =

⇒ lim min( )sin

xy y

x

x→− +

0

2 4 11 =

→

limsin

x

x

x0

7

Now, | sin | | |x x< , where x → 0

∴ limx→0

[a value slightly less than 7]

= 6

TEST Edge Evaluation of limits of the logarithmicfunction, exponential function related questions areasked. To solve such types of questions, studentsare advised to understand the concept of limit andalso acquainted yourself with properties of the limitsuch as

lim

,

,

,

,

n

na

a

a

a

a

→ ∞=

∞ >=

− < <

if

if

if

if

1

1 1

0 1 1

doesn,t exist ≤ −

1

PRACTICE SET 2 101

a

r n· = d

51. (a,c) IdeaQ f x x( ) [ ]= is a greatest integer function. If

n x n≤ < + 1, then [ ] ,x n= such as [2. 3] = 2. [9.2]= 9 and apply the concept of limit to get therequired solution.

lim [ ] lim [ ] lim ( )x x x

x x→ → →

− = − = − = −0 0 0

0 1 1.

We have x 2 0> and lim( )

x

f x

x→=

0 22

∴ f x( ) must be a positive quantity.

Also, x 2 0→ as x → 0 and limit is finite.

∴ f x( ) should also approach to zero as x → 0

lim [ ( )]x

f x→

=0

0

Now, lim( )

lim( )

x x

f x

xx

f x

x→ →+ +

= ⋅

=0 0

20

lim( )

lim( )

x x

f x

xx

f x

x→ →− −

= ⋅

= −0 0

21

∴ lim( )

x

f x

x→

0

does not exist.

TEST Edge Evaluation of limit of algebraic function,modulus function, or related questions are asked.To solve such types of questions, students areadvised to understand the concepts of the limit andalso acquainted yourself with the properties of limitsuch as if

lim ( ) lim ( ) ,x a x a

f x A g x B→ →

= > =0 and

then lim [ ( )] ( )

x a

g x Bf x A→

= .

52. (a,c) Idea Q Let the point of intersection be ( , ),h k

then compare the both equations and solvethem in ( , )h k to get required solution.

Let the point of intersection of E1 and E 2 be ( , )h k

∴ h

ak

2

21+ =2 and h

k

a

22

21+ =

⇒ h

ak

2

221= − and 1 2

2

2− =h

k

a

⇒ h

ka

2

21−= 2 and a

k

h

22

21=

−

∴ h

k

k

h

2

2

2

21 1−=

−⇒ h h k k2 2 2 21 1( ) ( )− = −⇒ h h k k2 4 2 4− = −⇒ h k h k2 2 4 4 0− − + =⇒ h k h k2 2 4 4 0− − − =( )

⇒ ( ) ( )( )h k h k h k2 2 2 2 2 2 0− − − + =⇒ ( )[ ( )]h k h k2 2 2 21 0− − + =⇒ h k h k2 2 2 2 1= + =or

⇒ x y x y2 2 2 2 1= + =or

⇒ x y y= ± + =or x2 2 1

TEST Edge Locus of straight line, parabola, and circlerelated questions are asked. To solve such types ofquestions, students are advised to understand theconcepts of locus and also acquainted yourself withthe properties of straight line, parabola, and circlesuch as, the two lines a x b y c1 1 1 0+ + = and

a x b y c2 2 2 0+ + = are intersecting, ifa

a

b

b1

2

1

2

≠ i e. . , if

they are neither coincident nor parallel.

53. (b,c) Idea To sovle this problem first of all find the nth

term of the given series, and use method ofdifference, then apply the concept ofsummation of series.

Tr

r =+

−tan 1

22

1

Sr r

r r

r rn

r

n

r

n

=+ +

= + −

+ +−

=

−

=∑ ∑tan tan

(1

12

1

1

2

2 1

2

1 2)

= + −−

=

−∑ tan ( ) tan1

1

12r

n

r r

Sn = −− −[tan tan ]1 13 1 + −− −[tan tan ]1 14 2+ −− −[tan tan ]1 15 3

+ ................. [tan ( ) tan ]− −+ −1 12n n

= + + + − −− − − −tan tan1 1 1 12 1 2 1( ) [ ] tan tann n

Sn n

n nn = +

+ +

−tan 12

2

3 7

9 10

S n

n n

nn

=+

+ +

→∞

−lim tan 1

2

37

19 10

= −tan ( )1 3

or cot−

1 1

3

TEST Edge Conceptual mixing of trigonometricfunction, algebraic function and series basedquestions are asked. To solve such types ofquestions, students are advised to understand theconcepts of series and also acquainted yourself withthe properties of the functions, such as if a is the firstterm, d the common difference and r the commonratio then, sum of first n terms ofarithmetic-geometric progression (AGP) then

Sd

r

dr r

r

a n d r

rn

n n

=−

+−−

−+ −

−

−

( )

( )

( )

[ ( ) ]

( )1

1

1

1

1

1

2

54. (a,c) IdeaQ If ax bx c2 0+ + = has roots α and β then

α β+ = − b a/ , αβ = c a/ and if sum of roots be Sand their product be Pthen quadratic equationx Sx P2 0− + =

Given that u and v are roots of equation

x px q2 0+ + =So, u v p+ = −

uv q=


PRACTICE SET 2 103

a. Now, equation whose roots are1

vand

1

u

⇒ xu v

xuv

2 1 1 10− +

+ =

xu v

uvx

uv

2 10− + + =

uvx u v x2 1 0− + + =( )

⇒ qx px2 1 0+ + +b. Now equation whose roots are u2 and v 2

So, equation will be

x u v x u v2 2 2 2 2 0− + + =( )

So, equation will be

x p q x q2 2 22 0− − + =( )

Q u v u v uv2 2 2 2+ = + −( ) p q2 2−and u v q2 2 2=

TEST Edge Common roots, greatest and least valuesof a quadratic expression, nature of roots of aquadratic equation with respect to one or two realnumbers related questions are asked. To solve suchtypes of questions, students are advised tounderstand the concept of quadratic equation andexpression such as a polynomial equation f x( ) = 0has exactly r roots equal to α iff f f f n( ) ( ) ( ) ... ( )α α α α= ′ = ′′ =− 1 0 and f n( )α ≠ 0.

55. (a,b,c) Idea Q If three vectors a i j k= + +a a a1 2 3$ $ $ ,

b i j k= + +b b b1 2 3$ $ $ and c i j k= + +c c c1 2 3

$ $ $

are coplanar then [ ]a b c = 0 ora a a

b b b

c c c

1 2 3

1 2 3

1 2 3

0= and if two vectors p and q

are perpendicular to each other, thenp q⋅ = 0.

Given vectors are coplanar

⇒a b c

b c a

c a b

= 0

⇒ a b c ab c3 3 3 3 0+ + − =⇒ ( ) ( )a b c a b c ab bc ca+ + + + − − − =2 2 2 0

= + + + + − − − =1

22 2 2 2 2 2 02 2 2( )( )a b c a b c ab bc ca

⇒ 1

202 2 2( )[( ) ( ) ( ) ]a b c a b b c c a+ + − + − + − =

⇒ a b c+ + = 0

or a b c= =But a b≠∴ a b c+ + = 0

Now, α ⋅ = + + ⋅ +v i j k i j+k( $ $ $ ) ($ $ $ )a b c

⇒ a b c+ + = 0

β.v = + + =a b c 0

λ ⋅ = + + =v a b c 0

TEST Edge Collinear vectors linearly dependent orindependent vector and geometrical application ofdot product of two vectors related questions areasked. To solve such types of questions, studentsare advised to understand the concept of typesvector and dot product of two vectors such as ifa i j k= + +a a a1 2 3

$ $ $ , b i j k= + +b b b1 2 3$ $ $ then by

Cauchy-schwarz inequality ( ) | | | |a b a b⋅ ≤ 2 2.

56. (5) Idea Q If z a ib= + is a complex number then

z a ib= − and z z a+ = 2 , z z ib− = 2 and| | ,| | ( ) ( )z zz z a z a z a2 2= − = − − . Now, applythe concept of minimum to get the requiredsolution.

p z z z z z i z i= + − − + − +( )( ) ( )( )3 3 6 6

= − + + + − +3 3 9 6 36z z z z z z i( ) ( )

= + − + + +3 3 2 9 6 2 362 2( ) ( ) ( )x y x i y i(

)

let z

then

= += −

+ =− =

x iy

z x iy

z z x

z z iy

2

2

= + − + − +3 6 9 12 362 2( )x y x y

= + − − +3 2 4 152 2[ ]x y x y

= − + − +3 1 2 102 2[( ) ( ) ]x y

For minimum value of p, x =1, y = 2

Minimum value of p = =3 10 30( ) ⇒ q

6

30

65= =

TEST Edge Principal value of arg of complexnumber, properties of modulus and properties ofargument related questions are asked. To solve suchtypes of questions, students are advised tounderstand the concept of complex number such as

| | | | | | | | | |z z z z z z1 2 1 2 1 2− ≤ + ≤ + . Thus, | | | |z z1 2+ isthe greatest possible value of | |z z1 2+ and| | | |z z1 2− is least possible for value of | |z z1 2+ .

57. (8) Idea Q x a= cos θ, y b= sinθ are the parametric

equations of the ellipsex

a

y

b

2

2

2

2 1+ = , where

Parameter 0 2≤ <θ π.

x = ∈cos [ , ]θ, θ π0 and x ∈ −[ , ]1 1

also apply the concept of maxima and minima.

Let x − =4 2 cos θ ⇒ x = +2 4cos θand y = 3 sinθ

Now, Ex y= +

2 2

4 9

= + +( cos )sin

2 4

4

22θ θ

= + + +4 16 16 4

4

2 2cos cos sinθ θ θ

= + = +20 16

45 4

coscos

θ θ

Now, Emax = + =5 4 9 [Q cos [ , ]θ ∈ − 11 ]

⇒ Emin = − =5 4 1

⇒ E Emax min− = − =9 1 8

AdvancedTest RIDER

TEST Edge Properties of an ellipse, eccentricity of theellipse, auxiliary circle related questions are asked.

To solve such types of questions, students areadvised to understand the concepts of ellipse suchas the equations of chord joining the pointsP a b( cos , sin )θ θ and Q a b( cos , sinθ θ )2 2 is

x

a

y

bcos sin cos

θ θ θ θ θ θ1 2 1 2 1 2

2 2 2

+

++

=−

58. (4) IdeaQVector area of a ∆ABC, when a, b, care the

position vectors of A B C, , respectively is area

( ) ( )∆ABC = × + × + ×1

2a b b c c a

Let OA a OB b OC c OD d= = = =, , ,

AB b a BC c b CD d c= − = − = −, ,

AD d a CA a c BD d b= − = − = −, ,

Then, AB CD BC AD CA BD× + × + ×= − × − + − × −( ) ( ) ( ) ( )b a d c c b d a

+ − × −( ) ( )a c d b

= × − × − × + ×b d b c a d a c

+ × − × − × + ×c d c a b d b a

+ × − × − × + ×a d a b c d c b

= × + × + ×2 2 2( ) ( ) ( )b a c b a c

Now, AB CD BC AD CA BD× + × + ×= × + × + ×2 2 2( ) ( ) ( )b a c b a c

= × + × ×2| |a b b c+c a

= ⋅ × + × + ×41

2| |a b b c c a

= 4 (area of ∆ABC)

TEST Edge Geometrical application of dot productor cross products of two vectors related questionsare asked. To solve such types of questions,students are advised to understand the concept ofproduct of two vectors such as the area of aparallelogram with an adjacent sides of a and b isgiven by |a × b|.

59. (2) Idea To solve this problem use the concept oftheory of equation and apply concept ofnumber of integral solution.

For an integer solution, it must lie between 1 and 2007(for otherwise all terms will be positive or negative).

Now, the terms will cancel in pairs if integer is takenas

2007 1

21004

+ =

⇒ K

502

1004

5022= =

TEST Edge Location of roots, logarithmicinequalities related questions are asked. To solvesuch types of questions, students are advised to

understand the concept of theory of equation suchas the condition that a quadratic functionf x y ax hxy by gx fy c( , ) = + + + + +2 22 2 2 may be

resolved into two linear factor is that

a h g

h b f

g f c

= 0

60. (5) Idea To solve this problem, use the concepts,image of the orthocentre of a triangle in any ofits sides always circumcentre on its lies.

Image of point P( , )7 2− in the line x = 2 is ( , – )− 3 2

Orthocentre of ∆ is ( , – )−3 2

To solve this problem, use the concept of the imageof point of intersection of altitude with circumcircle isthe orthocenter of triangle

∴ Absolute value of sum of coordinates = − −| |3 2

= 5

TEST Edge Image or reflection of a point in differentcases, and application of image or reflection relatedquestions are asked. To solve such types ofquestions, students are advised to understand theconcept of image or reflection of a points in 2D suchas the image of A x y( , )1 1 with respect to line mirrorax by c+ + =0 be B h k( , ) is given byh x

a

k y

b

ax by c

a b

−=

−=

− + ++

1 1 1 12 2

2( ).

104

a b2 2+

A x y( , )1 1

B h k( , )

ax by + c+ = 0

3 4 5 6 7

55

(–3, –2) (7, –2)x = 2

O

A

C

B

test rider advanced - m.media-amazon.com

Documents