1

PHYSICS 1. A light string passing over a smooth light pulley connects two blocks of masses 1m and 2m (vertically). If the

acceleration of the system is g/8 then the ratio of the masses is (A) 8 : 1 (B) 9 : 7 (C) 4 : 3 (D) 5 : 3

Solution (B):

1 2

1 2

m ma g

m m

1 2

1 2

m mgg

8 m m

1

2

m 9

m 7

2. If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

(A) 5 2, / 3 (B) 5 2, / 4 (C) 5 2, / 3 (D) 5 2, / 4

Solution (B):

2 2F 5 5 5 2 N. and 5

tan5

=1

/ 4.

3. A body of mass 3kg is acted on by a force which varies as shown in the graph below. The momentum

acquired is given by

(A) Zero (B) 5 N-s (C) 30 N-s (D) 50 N-s Solution (D):

momentum acquired = Area of force-time graph

1

(2) (10) 4 10 10 40 50 N-S2

4. A machine gun fires 20 bullets per second into a target. Each bullet weighs 150 gms and has a speed of

800 m/sec. Find the force necessary to hold the gun in position (A) 800 N (B) 1000 N (C) 1200 N (D) 2400 N Solution (D):

3F mnv 150 10 20 800 2400 N.

5. Two weights 1w and 2w are suspended from the ends of a light string passing over a smooth fixed pulley. If

the pulley is pulled up at an acceleration g, the tension in the string will be

(A) 1 2

1 2

4w w

w w (B) 1 2

1 2

2w w

w w (C) 1 2

1 2

w w

w w (D) 1 2

1 2

w w

2(w w )

Solution (A):

1 2

1 2

2m mT (g a)

(m m )

1 2

1 2

2m m (g g)

m m

1 2 1 2

1 2 1 2

4m m 4w wT g

m m w w

2

6. A body takes just twice the time as long to slide down a plane inclined at 30o to the horizontal as if the plane were frictionless. The coefficient of friction between the body and the plane is

(A) 3

4 (B) 3 (C)

4

3 (D)

3

4

Solution (A):

2 2

1 1 3tan 1 tan30 1

4n 2

7. A motorcyclist of mass m is to negotiate a curve of radius r with a speed v. The minimum value of the

coefficient of friction so that this negotiation may take place safely, is

(A) 2v rg (B) 2v

gr (C)

2

gr

v (D)

2

g

v r

Solution (B): Conceptual

8. A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table

surface is 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table is

(A) 20% (B) 25% (C) 35% (D) 15% Solution (A):

' 0.25 ll l l 20%

1 0.25 1 5

of l.

9. A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is applied on it. If

the coefficient of kinetic friction is 0.4 and 2g 9.8m / s , then the acceleration of the block is

(A) 20.26m / s (B) 20.39m / s (C) 20.69m / s (D) 20.88m / s

Solution (D): Net force = Applied force – Friction force

ma 24 mg 24 0.4 5 9.8 24 19.6 24.4a 0.88 m / s

5

10. When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts (A) In the backward direction on the front wheel and in the forward direction on the rear wheel (B) In the forward direction on the front wheel and in the backward direction on the rear wheel (C) In the backward direction on front and no friction on the rear wheels (D) In the forward direction on both front and the rear wheels Solution (A):

In cycling, the rear wheel moves by the force communicated to it by pedaling while front wheel moves by itself. So, while pedaling a bicycle, the force exerted by rear wheel on ground makes force of friction act on it in the forward direction (like walking). Front wheel moving by itself experience force of friction in backward direction (like rolling of a ball). [However, if pedaling is stopped both wheels move by themselves and so experience force of friction in backward direction].

11. A body of mass 10kg at rest is acted upon simultaneously by two forces 4 N and 3N at right angles to each

other. The kinetic energy of the body at the end of 10 sec is (A) 100 J (B) 300 J (C) 50 J (D) 125 J

Solution (D):

Net force on body 2 24 3 5N 2a F / m 5 / 10 1/ 2m/s

Kinetic energy = 21mv

2

21m(at) 125Joule

2

3

12. Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in the ratio (A) 4 : 1 (B) 1 : 4 (C) 2 : 1 (D) 1 : 2

Solution (C): 2F

U2k

1 2

2 1

U k

U k (if force are same) 1

2

U 3000 2

U 1500 1

13. The potential energy between two atoms in a molecule is given by 12 6

a bU(x)

x x ; where a and b are

positive constants and x is the distance between the atoms. The atom is in stable equilibrium when

(A) 611a

x5b

(B) 6a

x2b

(C) x 0 (D) 62a

xb

Solution (D):

Condition for stable equilibrium dU

F 0dx

12 6

d a b0

dx x x

13 712ax 6bx 0

13 7

12a 6b

x x 62a

xb

62a

xb

14. If the linear momentum is increased by 50%, the kinetic energy will increase by

(A) 50% (B) 100% (C) 125% (D) 25% Solution (C):

Let 1P P , 2 1P P 50% of 1P = 1 11

P 3PP

2 2

2E P

2 22 2 1

1 1 1

E P 3P /2 9

E P P 4

2E 2.25E 1 1E 1.25E

2 1 1E E 125% of E

i.e. kinetic energy will increase by 125%.

15. A force of ˆ ˆ ˆ2i 3j 4k N acts on a body for 4 second, produces a displacement of ˆ ˆ ˆ(3i 4j 5k)m. The

power used is (A) 9.5 W (B) 7.5 W (C) 6.5 W (D) 4.5 W

Solution (A): ˆ ˆ ˆ ˆ ˆ ˆF.s (2i 3 j 4k).(3i 4 j 5k)

Pt 4

389.5 W

4

16. A gun fires a bullet of mass 50 gm with a velocity of 130msec . Because of this the gun is pushed back

with a velocity of 11msec . The mass of the gun is

(A) 15 kg (B) 30 kg (C) 1.5 kg (D) 20 kg Solution (C):

B BG

G

m vm

v

350 10 301.5 kg

1

4

17. A ball of mass m falls vertically to the ground from a height h1 and rebound to a height 2h . The change in

momentum of the ball on striking the ground is

(A) 1 2mg(h h ) (B) 1 2m( 2gh 2gh )

(C) 1 2m 2g(h h ) (D) 1 2m 2g(h h )

Solution (B):

When ball falls vertically downward from height 1h its velocity 1 1v 2gh

and its velocity after collision 2 2v 2gh

Change in momentum

2 1 1 2P m(v v ) m( 2gh 2gh )

(because 1v and 2v are opposite in direction)

18. A bag (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets

caught in the bag. Then for the combined (bag + bullet) system

(A) Momentum is mvM

M m (B) Kinetic energy is

2mv

2

(C) Momentum is mv(M m)

M

(D) Kinetic energy is

2 2m v

2(M m)

Solution (D):

Initial momentum = mv Final momentum = (m M)V

By conservation of momentum mv (m M)V

Velocity of (bag + bullet) systemmv

VM m

Kinetic energy = 21(m M) V

2

=2

1 mv(m M)

2 M m

2 21 m v

2 M m

19. A weight lifter lifts 300 kg from the ground to a height of 2 meter in 3 second. The average power

generated by him is (A) 5880 watt (B) 4410 watt (C) 2205 watt (D) 1960 watt

Solution (D):

PWorkdone

Time =

mgh

t

300 9.8 21960 W

3

5

20. A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E. of colliding body before and after collision will be (A) 1 : 1 (B) 2 : 1 (C) 4 : 1 (D) 9 : 1

Solution (D):

K.E. of colliding body before collision 21mv

2

After collision its velocity becomes

1 2

1 2

(m m ) m vv v v

(m m ) 3m 3

K.E. after collision 21mv '

2

21 mv

2 9

Ratio of kinetic energy =

2

before2

after

1mv

K.E. 2 9 :1K.E. 1 mv

2 9

21. A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to

height h. The time taken by the stone to reach the ground is

(A) h

4g

(B) h

2g

(C) 2h

g (D)

g

h

Solution (B):

The velocity of balloon at height h, ghg gh

v 2 h8 4 2

When the stone released from this balloon, it will go upward with velocity v =gh

2 (Same as that of balloon). In

this condition time taken by stone to reach the ground

2

v 2ght 1 1

g v

gh / 2 2gh

1 1g gh / 4

2 gh h2

g g

22. A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which

reaches a height 4 h. The two stones cross each other after time

(A) h

8g (B) 8gh (C) 2gh (D)

h

2g

Solution (A): For first stone u 0 and

For second stone 2

2u4h u 8gh

2g

u 8gh

Now, 21

1h gt

2

22

1h 8ght gt

2

where, t =time to cross each other.

1 2h h h

2 21 1gt 8ght gt h

2 2

h ht

8g8gh

6

23. A projectile projected at an angle o30 from the horizontal has a range R. If the angle of projection at the

same initial velocity be o60 , then the range will be

(A) R (B) 2R (C) R / 2 (D) 2R Solution (A): For complementary angles of projection horizontal range is same.

24. A body of mass m is projected at an angle of o45 with the horizontal. If air resistance is negligible, then

total change in momentum when it strikes the ground is

(A) 2mv (B) 2 mv (C) mv (D) mv / 2

Solution (B):

Change in momentum= 2mvsin 2mv sin4

= 2 mv

25. A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest

height attained by it. The range of the projectile is (where g is acceleration due to gravity)

(A) 24v

5g (B)

2

4g

5v (C)

2v

g (D)

24v

5g

Solution (A):

R 2H given

We know R 4Hcot 1

cot2

From triangle we can say that 2

sin5

,1

cos5

Range of projectile 22v sin cos

Rg

22v 2 1

g 5 5 =

24v

5g.

26. A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass m at a point is

in horizontal direction, find the tangential force at this point in terms of tension T and mg.

(A) mg (B) 2 2mgT (mg)

T

(C) 2 2mg(mg) T

T (D) 2 2T

(mg) Tmg

Solution (B): When the acceleration of bob is horizontal, net vertical force on the bob will be zero. T cos θ – mg = 0 The tangential force at that instant is

= mg sin θ = mg 2 2 2mg1 cos θ = T (mg)

T

7

27. A body moves along an uneven surface with constant speed at all points. The normal reaction of the road on the body is:

(A) maximum at A (B) maximum at B

(C) minimum at C (D) the same at A, B & C Solution (A):

2 2mv mvN mg N mg

R r

r min at A N max at A

As by energy conservation RA < RC

NA is greatest among all. 28. A disc having radius r = 2m is placed on smooth horizontal surface velocity of point A and B is given as

shown in figure. Distance between point O and axis of rotation at that particular instant of time. (A) 1 m (B) 0.5 m (C) 0.25 m (D) 2 m Solution (B):

2.5 1.5

(2 y) (2 y)

3 1.5y 8 4y

3 1.5y 5 2.5y

4y = 2 ; y = 0.5 m 29. In the given figure the fixed pulley is smooth and light where as the movable pulley is rough and has some

mass. There is no slipping between the pulley and the string. Then magnitude of acceleration of the point ‘P’ on the pulley is(if block is moving with an acceleration ‘a’)

(A) a (B) 2

a

(C) 2a (D) 2

a

Solution (D):

Acceleration of pulley is a

2 and

a

2R

So Pa

a2

30. If cmL is the angular momentum about COM of a system of particles , cmr is the position vector of the

centre of mass relative to any point ‘P’ and p is the total momentum the angular momentum L of the

system relative to the point P is given by

(A) cmL L (B) cm cmL p r L

(C) cm cmL r p L (D) cm cmL L r p Solution (C):

cm cmL L r P

2.5 m/sec

1.5 m/sec

O

B

A

8

CHEMISTRY 31. A 2g sample of xenon react with fluorine. The mass of compound produced is 3.158 g. The empirical

formula of the compound is: (Xe 131, F 19)

(A) 2XeF (B) 4XeF (C) 5XeF (D) 6XeF

Solution (B):

Mass of Xe 2 Xe2

n 0.015131

Mass of F 1.158 F1.158

n 0.06019

Xe

F

n 1

n 4

Empirical formula 4XeF

32. 1.625g of an interhalogen compound xICl was subjected to electric discharge which spilt it into

constituents and brought to STP. It occupied 224 ml. The interhalogen compound is: (I 127, Cl 355)

(A) ICl (B) 3ICl (C) 5ICl (D) 7ICl

Solution (A):

a

2 2a a a

2 2

2ICl I Cl

2224a 10 mol

22400

No. of mole 21.62510

M.Mass

Molar mass 162.5 ICl

33. On passing 1.8 L of oxygen through ozonizer gives 1.6 L of a mixture of ozone and oxygen. What is the mole fraction of ozone in the resultant mixture?

(A) 0.15 (B) 0.25 (C) 0.40 (D) 0.50 Solution (B):

1.8

2 321.8 x x3

3O 2O

3O

2x 2 0.6n 0.4

3 3

2x

1.8 x 1.63

3O

0.4 0.25x

1.6

x1.8 1.6

3

x

0.2 x 0.63

34. A and B are two elements which form 2 3A B and 2AB . If 0.30 mole of 2 3A B weighs 31.8 g and 0.30 mole

of 2AB weighs 18.6 g. What are the atomic weights of A and B respectively:

(A) 18, 26 (B) 26, 18 (C) 9, 13 (D) 26,52 Solution (B):

31.8

2A 3B 1060.3

…(i)

18.6

A 2B 620.3

…(ii)

From equation (i) & (ii): B 18 A 26

9

35. In which of following case would the probability of finding an electron residing in a xyd orbital be zero?

(A) xy and yz plane (B) xy and xz planes (C) xz and yz planes (D) z-direction, yz and xz planes Solution (C):

36. 4000 Å photon is used to break the 2I molecule, then calculate the percentage of energy converted to the

KE of atoms if bond dissociation energy of 2I molecule is 246.5 kJ/mole.

(A) 12% (B) 8% (C) 26% (D) 17% Solution (D):

20hcE 5 10 kJ

x

Bond dissociation energy/moleculeA

246.5

N

% of energy converted to K.E. 20

A

246.55 10 100 17%

N

37. According to Bohr’s atomic theory, which of the following relation is incorrect?

(A) Frequency of revolution of the electron in an orbit

2

3

Z

n

(B) Coulombic force of attraction on the electron

3

4

Z

n

(C) Kinetic energy of electron

2

2

Z

n

(D) The product of velocity of electron and the principal quantum number 2Z

Solution (D): Consider Bohr’s Model

38. The wavelength of series limit for Lyman series for hydrogen would be: (A) 911.76Å (B) 819.23 Å (C) 616.07 Å (D) 576.25 Å

Solution (A):

2

2

1 1 1RZ Z 1

X 1

39. The set of quantum numbers which can explain the last electron in 63d configuration is:

(A) 1

n 3, 2,m 0,s2

(B) 1

n 3, 2,m 2,s2

(C) 1

n 3, 1,m 1,s2

(D) 1

n 3, 2,m 2,s2

Solution (B):

For 63d n 3 2l m 2

1

s2

y

x

10

40. Which one of the following metal ions has the smallest size in aqueous solution?

(A) Li (B) Na (C) K (D) Cs

Solution (D):

Less hydration of Cs

41. The set representing the correct order of first ionization potential is: (A) K Na Li (B) Be Mg Ca (C) B C N (D) Ge Si C

Solution (B):

2 2Be 1s ,2s 2 2 6 2Mg 1s ,2s ,2p ,3s Be Mg Ca

2 2 6 2 6 2Ca 1s ,2s ,2p ,3s ,3p ,4s

42. The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is: (A) O F N C (B) F O N C

(C) C N O F (D) O N F C

Solution (A):

3 4O 2p , F 2p 2 1N 2p , C 2p

Half filled orbital are more stable than partial field.

O F N C

43. The first ionization potential of Na, Mg, Al and Si follow the order: (A) Na Mg Al Si (B) Na Mg Al Si

(C) Na Mg Al Si (D) Na Mg Al Si

Solution (B): From left to right in periodic table ionization energy increases. soMg Na from Al electron should be

removed from P subshell, so it is less than Mg. 44. What is the group number and period number of an element having atomic number 34? (A) Group number 6, Period number 4 (B) Group number 16, Period number 4 (C) Group number 15, Period number 5 (D) Group number 14, Period number 3

Solution (B):

2 2 6 2 6 2 10 434 1s ,2s ,2p ,3s ,3p ,4s ,3d ,4p

2 2 6 2 6 10 2 41s ,2s ,2p ,3s ,3p ,3d ,4s ,4p

That is why group No. =16. Period no=4. 45. Which of the following pairs of molecules/ions are not likely to exist

(A) 22 2H ,He (B) 2

2 2H ,He (C) 22 2H ,He (D) 2

2 2H ,He

Solution (C): Species having zero or negative bond order do not exist. 2 02H (1 1 2 0) 1s

Bond order =0 2 * 2

2He (2 2 4) 1s , 1s

Bond order b aN N 2 20

2 2

So, both 22H and 2He do not exist.

46. Which element form p p multiple bonds with itself and with carbon and oxygen?

(A) P, As (B) N, As (C) N, P (D) N Solution (D):

Fact.

11

47. Molecular shapes of 4SF , 4CF and 4XeF are:

(A) Same with 2, 0 and 1 lone pairs of electrons respectively (B) The same with 1, 1 and 1 lone pairs of electrons respectively (C) Different with 0, 1 and 2 lone pairs of electrons respectively (D) Different with 1, 0 and 2 lone pairs of electrons respectively

Solution (D):

4: SF One lone pair

4CF 0 lone pair

4: XeF 2 lone pair

48. Which among the following factors is the most important in making fluorine the strongest oxidizing agent?

(A) Electron affinity (B) Ionisation enthalpy (C) Hydration enthalpy (D) Bond dissociation energy Solution (C):

Due to small size of fluorine, it has high hydration energy which makes it strongest oxidizing agent.

49. The bond energies in NO, NO and NO follow the order:

(A) NO NO NO (B) NO NO NO

(C) NO NO NO (D) NO NO NO

Solution (A):

N O B.O 3 NO B.O 2.5

NO BO 2

50. van der Waal’s equation for 4CH at low pressure is:

(A) PV RT Pb (B) a

PV RTV

(C) a

PV RTv

(D) PV RT Pb

Solution (B): At low pressure Van der Waal’s constant b=0

2

aP (V b) RT

V

a

PV RTV

51. Four one litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same

room temperature and pressure. The ratio of total number of atoms of these gases present in different flasks would be:

(A) 1:1:1:1 (B) 1:2:2:3 (C) 2:1:2:3 (D) 3:2:2:1 Solution (C):

All the gases have same mole

2 2 3H :He,O :O

2 : 1 : 2 : 3

12

52. For the process

2 2H O( ) H O(g)l

At oT 100 C and1atm, the correct choice is:

(A) system surroundingS 0 and S 0 (B) system surrounding0 and 0

(C) system surroundingS 0 and S 0 (D) system surroundingS 0 and S 0

Solution (B):

Entropy of system increases because 2 2H O( ) H O(g)l because system absorbs heat entropy of

surrounding decreases.

53. Equal amounts of 2N and Ar have their total translational kinetic energy in the ratio 5:1. What would be the

ratio of temperature of 2N to that of Ar?

(A) 10:7 (B) 7:2 (C) 7:4 (D) 9:4 Solution (B):

3

K.E n. RT2

2N 1

w(K.E) RT

28

Ar 2w

(K.E) RT40

2N 1

Ar 2

(K.E) T5 40

(K.E) 1 28 T

2

1

T 2

T 7

2T : T 7 : 2

54. Which of the following represent heat of formation.

(A) (diamond) 2 2C O (g) CO (g) (B) 2 22CO(g) O (g) 2CO (g)

(C) 2 21 1H (g) F (g) HF(g)

2 2 (D) 2 2 3N (g) 3H (g) 2NH (g)

Solution (C):

ofH is defined as energy involved when one mole of compound is formed from its element in standard

state.

55. AB, 2 2A and B are diatomic molecules. If the bond enthalpies of 2A , AB and 2B are in ratio 1:1:0.5 and

the enthalpy of formation of AB from 2 2A and B is 1100kJmol , what is the bond enthalpy of

1

2A in kJmol ?

(A) 400 (B) 200 (C) 100 (D) 300 Solution (A):

a/2aa

2 21 1A B AB

2 2

of

1 1 1100kJ H a a a a

2 4 4

a 400

13

2 2ClH C C C CH Cl

Br Br

3H C C C COOH

3H C 3CH

Cl

O

3 3CH C CH C OCH

Cl O

56. Latent heat of vaporization of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy ( E) of 3 moles of liquid at same temperature?

(A) 30 kcal (B) 54 kcal (C) 27.0 kcal (D) 50 kcal

Solution (C):

gH E n RT

Given, H 30 kcal for 3 mole gn 3

Because, liquid vapour

330 E 3 2 500 10 E 27kcal

57. When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse through out the

entire room. The incorrect fact about the process is:

(A) G ve (B) H 0 (C) S ve (D) S ve

Solution (C): Because process in spontaneous ( S ve)

58. The Correct IUPAC name of the following compound is:

(A) 1,1,1,5,5,5-Hexachloro-2,4-diethylpentane (B) 3,5-Hexachlorodimethylheptane (C) 3,3,3,5,5,5-Hexachloromethylheptane (D) 3,5-B is (trichloromethyl)heptanes Solution (D):

59. Which of the given is incorrect IUPAC names: (A) 2,3-Dichlorocyclohex-1-ene (B) 2,5-Dimethyloct-4-ene (C) 4-bromo-1-chloropentane (D) 6-Ethyl-2,3-dimethylnonane Solution (A):

60. Which of the following is correct IUPAC names? (I) Methyl 3-chloro-2-butenoate (II) 2,3-Dimethylpentanoyl chloride (III) 3-Metylbut-2-enoic acid (IV) 2,3-Dibromo-1,4-dichloro-2-butene

(A) I, II only (B) II, III only (C) III, IV only (D) All of these

Solution (D):

ClCl Cl Cl ClCl

14

MATHEMATICS

61. If 1 1x ,y and 2 2x ,y are the solution of the system of equation 225 64log x log y 4 ,

x ylog 225 log 64 1, then the value of 30 1 1 2 2log x y x y is

(A) 1 (B) 12 (C) 0 (D) none of these Solution (B):

225 64log x log y 4 & x ylog 225 log 64 1

Let 225 64log x a & log y b

a b 4 …(i)

& 1 1

1a b

…(ii)

From (i) & (ii):

1 11

a 4 a

4 a a1

a 4 a

24 2a 4a a 2a 6a 4 0

Roots of the above equation are 225 1 225 2log x & log x

225 1 225 2log x log x 6

225 1 2log x x 6

6

1 2x x 225 …(iii)

Similarly from (i) & (ii):

1 11

4 b b

b 4 b1

4 b b

22b 4 4b b 2b 2b 4 0

Roots of the above equation are 64 1 64 2log y & log y

64 1 64 2log y log y 2

64 1 2log y y 2 2

1 2y y 64 …(iv)

From (iii) & (iv):

6 2

1 2 1 2x x y y 225 64

6 2

2 61 1 2 2x y x y 15 2

12 121 1 2 2x y x y 15 2

121 1 2 2x y x y 30

30 1 1 2 2log x y x y 12 .

62. If 1 2 3 10, , ,......, are the roots of the equation 10 9ax bx cx d 0 then the value of

1 2 3 10

1 2 3 10

.....

....... is

(A) d

b (B)

d

b (C)

b

d (D)

b

d

Solution (C):

1 2 10, ,....., are roots of 10 9ax bx cx d 0

1 2 10b

...a

& 1 2 10d

. .......a

15

1 2 10

1 2 10

b / a.... b

. ..... d / a d

63. If

2

1/2 2x 6x 9

log log x 1 ,2 x 1

then x lies in the interval

(A) 1, 1 2 2 (B) 1 2 2, 2

(C) 1, (D) None of these

Solution (A):

The log function are defined if

2x 6x 90

2 x 1and x + 1 > 0

2x 3

02 x 1

& x + 1 > 0 x > –1

Now the inequality is

but

64. If the quadratic equation, 2ax bx c 0, a, b, c R, a 0 and 2x 4x 5 0 have a common root then

a, b, c must satisfy the relation: (A) a > b > c (B) a < b < c

(C) a k, b 4k, c 5k, k R, k 0 (D) 2b 4ac is positive

Solution (C):

2x 4x 5 0have imaginary roots and 2ax bx c 0 also have imaginary roots.

2b 4ac 0 and a k, b 4k & c 5k, k R, k 0

65. If 2 1sin 2cos 0

4, then the general value of is

(A)

n3

(B)

2n3

(C)

2n6

(D)

n6

Solution (B):

2 1sin 2cos 0

4 2 1

1 cos 2cos2 04

2 5cos 2cos 0

4

52 4 4.1.

4cos

2

2 9

cos2

2 3

cos2

1 5

cos or cos2 2

(rejected)

1

cos2

cos cos

3

2n

3

16

66. General solution of the equation 3 3 1sin cos cos sin

4 is

(A)

n8

(B)

n8

(C)

n

2 8 (D)

n

2 8

Solution (C):

3 3 1sin cos cos sin

4

2 2 1sin cos sin cos

4

2 2 1sin cos cos sin

4

1 1

sin2 cos22 4

1 1

sin44 4

sin4 1

4 2n2

n

2 8

67. If 2 1sin

4, then the general value of is

(A)

n

2n 16

(B)

nn1

2 6 (C)

n

6 (D)

2n

6

Solution (C):

2 1sin

4

22 1

sin2

2 2sin sin6

n

6

68. STATEMENT I: If two sides of a triangle are 2 and 3, then its area cannot exceed 3.

STATEMENT II: Area of a triangle 1

bc sinA2

and sinA 1.

(A) STATEMENT I is True, STATEMENT II is True; STATEMENT II is a correct explanation for STATEMENT I.

(B) STATEMENT I is True, STATEMENT II is True; STATEMENT II is NOT a correct explanation for STATEMENT I.

(C) STATEMENT I is True, STATEMENT II is False. (D) STATEMENT I is False, STATEMENT II is True. Solution (A): Let b 2,c 3

1

bc sinA2

1

2.3sinA2

3sinA

sinA 1 3sinA 3

can not be greater than 3.

17

69. If the radius of the circum-circle of an isosceles triangle ABC is equal to AB AC , then angle A is

(A)

6 (B)

3 (C)

2 (D)

2

3

Solution (D):

a b

2RsinA sinB

b

2ACsinB

, ( R AB AC )

b

2bsinB

1

sinB2 oB 30

oB C 30 , (since ABC is an isosceles triangle with AB AC )

o o o oA 180 30 30 120 2 / 3

70. If 1 1 1

4 x 2 then x

(A) 4,2 (B)

1 1, ,

4 2 (C) , 4 2, (D) none of these

Solution (C):

1 1 1

4 x 2

1 1 1 1 10 or 0 or 0

4 x x x 2

4 x or x or x 2 x , 4 2,

71. If 2xx2 .5 then x

(A) ,0 1, (B) , 1 0, (C) , 1 1, (D) none of these

Solution (B):

2xx2 0.5

2xx 1

22

2x x2 2 2x x

2x x 0 x x 1 0 x 1 or x 0

x , 1 0,

72. If 2x sin 0 where ,2 then x, where [ ] denotes greatest integer

(A) , (B) 0, (C) ,0 (D) none of these

Solution (D):

Given 2 sin 1,0

sin 1, thus 2x sin 0 2x 1 0

2x 1 0 x .

73. If and are roots of 2ax bx 1 0 then the quadratic whose roots are 3 2a b 1 and 3 2a b 1

is

(A) 2ax b 2a x a b 1 0 (B) 2ax b 2a x a b 1 0

(C) 2ax b 2a x a b 1 0 (D) 2ax b 2a x a b 1 0

18

Solution (A):

Since and are roots of 2ax bx 1 0 therefore

b

a &

1

a. Also 2a b 1 0

2a b 1 0 .

3 2 2a b 1 a b 1 1 1 1

3 2 2a b 1 a b 1 1 1 1

Hence product of roots 1 1 1 1 b

1a a

1 b a

a

Sum of roots 2

b2

a

b 2a

a

Hence the required equation is:

2 b 2a a b 1x x 0

a a

2ax b 2a x a b 1 0 .

74. If 3 3 3a b c 3abc 0 , where a,b,c R & a b c 0 . If quadratic equation 2ax bx c 0 , has two

real roots & , then the value of

2 2 2 2

2 2

1 1 1 1

1 11 1

3 3

2 2

1 11 1

1 1 1 1

(A)

3 3 2 2

2 2

1 1 1 1 1 1

1 1 1 1 (B) 0

(C)

3 3

2 21 1 1 1 (D)

2 2

1

1 1 1 1

Solution (B):

3 3 3a b c 3abc 0

a b c 0 or a b c

As a b c 0 (given), therefore a b c 0 .

From observation we see that x 1satisfies the equation 2ax bx c 0 .

1 or 1

For 1 or 1

2 2 2 2

2 2

1 1 1 1

1 11 1

3 3

2 2

1 11 10

1 1 1 1

75. If ABC is right angled at C and sinA & cosB are roots of the equation 2p q r 0 , then

(A) 2q 4pr (B) 2p 4qr (C) 2 2q p 2pr (D) 2 2p q 2qr

Solution (A):

As ABC is right angled at C therefore

A B2

19

A B2

sinA cosB . As sinA & cosB are roots of the equation 2p q r 0 and sinA cosB ,

therefore roots of the equation 2p q r 0 are equal, and hence discriminant should be zero.

2q 4pr 0 2q 4pr .

76. If 2 2 2 2 2 2y log 3.log 12.log 48.log 192 16 log 12.log 48 10 , then value of y is

(A) 2 (B) 6 (C) 10 (D) 14 Solution (B):

2 2 2 2y log 3.log 12.log 48.log 192 16 2 2log 12 log 48 10

2 2 2 2 2 2 2y log 3 log 3 log 4 log 3 log 16 log 3 log 64 16

2 2 2 2log 3 log 4 log 3 log 16 10

Let 2log 3 x y x x 2 x 4 x 6 16 x 2 x 4 10

y x x 6 x 2 x 4 16 x 2 x 4 10

2 2 2y x 6x x 6x 8 16 x 6x 8 10

Let 2x 6x t

y t t 8 16 t 8 10 2y t 8t 16 t 8 10

2

y t 4 t 8 10 y t 4 t 8 10 y 6

77.

2

2

3x 7x 82

x 1 then

(A) x (B) x [–3, –2] (C) x 1, 6 (D) x 0, 10

Solution (C): Domain: x R

78. Number of solutions of the equation tanx sec x 2cosx lying in the interval [0, 2] is

(A) 0 (B) 1 (C) 2 (D) 3 Solution (C): The given equation can be written as

1 + sin x = 2 cos2x = 2(1- sin2x) 2 sin2x + sin x – 1 = 0 (1 + sin x)(2 sin x – 1) = 0

sin x = 1 or 1/2

Now sin x = 1 tan x and sec x not defined. Sin x = 1/2 x= /6 or 5/6.

The required number of solution is 2.

20

79. If no solution of 32siny 8sin x a lies on the line y = 3x, then

(A) a , 6 6, (B) a 6, 6

(C) a 6, 6 (D) None of these

Solution (A): 2siny + 8sin3x = a

No solution, if a , 6 6,

80. If x and y are solutions of the equation 3

sinx siny4

and 3

cos xcos y4

, then some of the value of x, y

(in digress) is

(A) ox 60 , oy 45 (B) o ox 30 , y 45

(C) o ox 30 , y 60 (D) o ox 60 , y 60

Solution (C): Adding and subtracting the equations

3

cos x y 0 and cos x y2

x y 2n 1 and x y 2n2 6

Now o o o ox y 90 , x y 30 x 60 , y 30

o o o ox y 90 , x y 30 x 30 , y 60

o o o ox y 90 , x y 30 x 30 , y 60

o o o ox y 90 , x y 30 x 30 , y 60

o o o ox y 90 , x y 30 x 60 , y 30

81. In ABC , if 2

cosA sinA 0cosB sinB

, then a b

c is equal to

(A) (B) 1 (C) (D)

Solution (A):

Given

82. In triangle ABC, s a 1 s b 1 s c 1

, ,8 12 24

then b

(A) 16 (B) 20 (C) 24 (D) 28

21

22 2

cos sin cos sin 2A A B B

cos sin 2A B A B

cos A B 1;sin A B 1

;2 2

A B A B C

2a b

c

21

Solution (A):

1 2 3

1 1 1 1

r r r r

2 1 3b r r r r

83. The value of a , for which one root of the equation is smaller than 1 and the

other greater than 2 is

(A) (B) (C) (D)

Solution (A):

(i)

--- (1)

(ii)

--- (2)

and

--- (3)

Using (1) , (2) & (3) The common condition is

84. The product of the roots of equation 4

log x103

10x

x is

(A) 3 (B) 4 (C) 310 (D) 410

Solution (C): Taking logarithm to the base 10 on both the sides of given equation:

2

10 10log x 4 3log x

2

10 10log x 3log x 4 0

10 1 10 2log x log x 3

Product of roots 10 1 2log x x 3

Hence 31 2x x 10

85. If 3125 0.2log (x 2) log (x 2) 6, then solution set of x is

(A) (1, ) (B) (0, 8) (C) (127, ) (D) (512, )

Solution (C):

3

125 0.2log x 2 log x 2 6

5 1/5

log x 2 log x 2 6

5

2 log x 2 6

5

log x 2 3

3x 2 5 x 127

25 2 4 0a x ax a

5,24a20

,3

a

5,a ,a

0D 24 4( 5)( 4) 0a a a

20 209 20 0 ,

9 9a a a

( 5) (1) 0; ( 5) (2) 0a f a f

( 5)( 5 2 4) 0a a a a

5 (5, )a a

( 5) ( 5).4 4 4 0a a a a

( 5)( 24) 0 5 24a a a

(5,24)a

(5,24)a

22

86. If 3

18

x then x

(A) x R (B) 1

, 0,2

(C) 0, (D) x

Solution (A):

31

8x

1

2x

1 1

2 0 or 0x x

1

x or x 02

1x , 0,

2

87. Let 2

f x x x 2 8 , where x denotes the greatest integer less than or equal to x, then

(A) f x 0only for one real values of x (B) f x 0 only for two real values of x

(C) f x 0only for three real values of x (D) f x 0 for infinite values of x

Solution

(D):

2f x x x 2 8

2

f x x x 2 8

2

f x x x 6 x 3 x 2

f x 0

x 3 0 or x 2 0 x 3 or x 2

x 3, 2 2,3

f x 0 for infinite real values of x.

88. If 6 6 2cos sin k sin 2 1 0, / 2 , then k is

(A) 3

4 (B)

1

4 (C)

1

3 (D)

1

8

Solution (A): The given condition can be written

3

2 2 2 2 2 2 2cos sin 3sin cos cos sin k sin 2 1

2 23sin 2 k sin 2 0

4

Showing that 3

k4

.

89. The value of 2 2cos 10 cos10 cos50 cos 50

(A) 4

5 (B)

1

3 (C)

3

4 (D) 3

23

Solution (C): 2 2cos 10 cos10 cos50 cos 50

1

1 cos20 cos60 cos40 1 cos1002

1 1 1 3 31 cos20 cos40 1 cos80 cos20 2cos60 cos20

2 2 2 2 4

90. If sin3x

f(x) ,x nsinx

then the range of values of f(x) for real values of x is

(A) [1, 3] (B) (, 1) (C) (3, ) (D) [1, 3) Solution (D):

3 – 4 sin2x = y

2 3 ysin x

4. But 0 < sin2x 1, ( sin x = 0 x = n )

3 y

0 14

or 0 < 3 – y 4.