mt 3 2010 2

7/28/2019 mt 3 2010 2

1/27

PHYSICS

IIT ians PACE Education Pvt. Ltd. : Andheri/Borivali/Dadar/Chembur/Thane/Churchgate/Nerul/Powai/Mulund

PART - I: PHYSICS

SECTION-I

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and

(D) for its answer, out which ONLY ONE is correct.

1. Force acting on a particle of mass m varies with time t as shown. Find work done by F on the

particle in time interval t = 0 to t = T. The particle starts from rest.

tT

F0

F

(a)2 2

0F T

8m(b)

2 2

0F T

4m(c)

2 2

0F T

2m(d) none of the above

2. A particle is projected with a speed 20 2 m/s at 45o with horizontal. After 1second, find tangential

acceleration, normal acceleration of particle, radius of curvature of trajectory at that point.

(g = 10 m/s2)

(a) 2 24 5 m / s , 2 5 m / s , 25 5 m (b) 2 25 m / s , 4 5 m / s , 20 5 m

(c) 2 22 5 m / s , 4 5 m / s , 25 5 m (d) none of the above

3. Bob of the pendulum AB is given initial velocity of 3gL in horizontal direction at

bottommost point . The maximum height reached by the bob from starting point if

AB is a massless rod is h1. The maximum height reached by bob from starting point

if AB is massless string is h2. The length of the pendulum is L Find ratio of1

2

h

h

.

(a)80

81(b)

81

80(c)

83

80(d) none of the above

4. A ball A, moving with kinetic energy makes a head-on elastic collision with stationary ball B of

mass n times that of A. Find maximum potential energy stored in system A + B during collision.

(a) zero (b)(n 1)

n 2

(c)

n

n 1

(d) none of above

5. Potential energy function along positive x axis is given by:

bU(x) axx

, (only for x > 0) where a and b are constants. It is known that system has only one

stable equilibrium configuration. Choose possible values of a and b out of the 4 given values.(a) a = 1, b = 2 (b) a = 1, b = -2 (c) a = -1, b = 2 (d) a = -1, b = -2

6. A, B, C and D are 4 identical spheres. A is moving, while B, C and D are at rest. A collides with Band then with C and comes to rest. B collides with D and comes to rest. Final velocities of C and D

are vC and vD, (vC > vD). All collisions are elastic. Find magnitude of initial velocity of A.

(a)2 2

C Dv v (b) vC + vD (c) vC - vD (d) none of the above

u 3gL

B

L

A

7/28/2019 mt 3 2010 2

2/27

PHYSICS


SECTION-II

This section contains 5 questions. The answer to each of the questions is a single digit integer,

ranging from 0 to 9.

7. A particle moves from origin O to point Palong a straight line2

xy . The work done by force

F

y i on particle for displacement OP

is3

K. Find K

(Force is in newtons, Displacement is in metres, work done is in joules

xy

2

2mx

Py

O

8. A body of mass1

6kg slides down a fixed inclined plane of angle 53

with horizontal. Coefficient of

friction between the body & inclined plane is 1. The rate at which mechanical energy of the body is

dissipated at any time tis kt. Find k. ( Take g = 10m/s2)

9. Minimum force of 500N is required to move 100 kgmass resting on a horizontal floor. is

coefficient of friction between body & floor. Find the value of2

1

. Take 210 m s .

10. A block, kept on top of a vertical spring, is allowed to fall. It compresses the spring by 5cm . When

the spring is cut into 2 equal parts & block is allowed to fall from same position as earlier, each

spring is compressed by 5cm.Natural length of the original spring is 110n

metres. Find n

11. A particle tied to a string attempting to perform vertical circular motion of radius12

11R m deviates

from circular path when modulus value of its radial acceleration is equal to its tangential

acceleration. In the entire motion, find the maximum height attained by particle above bottom most

point. (Take 2 1.5 )

7/28/2019 mt 3 2010 2

3/27

PHYSICS


SECTION-III

This section contains 2 groups of questions. Each group has 3 multiple choice question based on a

paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE

is correct.

Paragraph for Question Nos. 12 to 14

Comprehension # 1

2 blocks A & B having masses m1 = 1kg and m2 = 4kg are arranged as shown. The pulleys P

and Q are light and frictionless. All blocks are resting on horizontal floor and pulleys are held such

that strings remain just taut. At t = 0, a force F = 30 t starts acting on pulley P along vertically

upward direction as shown in fig. ( Take g = 10m/s2)

mA = 1kgBmB = 4kg

F = 30 t

A

P

Q

12. Find the time lag between the time instants that the 2 blocks leave ground

(a) 1s (b) 2s (c) 1.5s (d) None of the above

13. Find the velocity of A when B looses contact with ground

(a) 5 m/s (b) 6 m/s (c) 7 m/s (d) None of the above

14. Find the height through which A gets lifted upto the instant B loses contact with ground .

(a) 2m (b)4

m3

(c)5

m3

(d) none of the above


Comprehension # 2There is a boat of mass 250 kg standing in still water. 2 persons A of mass 50 kg and C of

mass 75 kg are standing on the edge of the boat. Both A and B can jump with a velocity 4 m/s

relative to the boat measured after the jump. The system is initially at rest. Ignore the friction effectsof water.

m2 = 75 kgm1 = 50 kg

Boat of mass 250 kg

7/28/2019 mt 3 2010 2

4/27

PHYSICS


15. The final speed of boat if both m1 and m2 jump simultaneously off the boat is

(a)5

4m/s (b)

4

3m/s (c) 1 m/s (d) none of above

16. The final speed of boat if m2 jumps first followed by m1 is

(a)23

17

m/s (b)21

19

m/s (c)22

15

m/s (d) none of above

17. Final speed of boat if m1 jumps first followed by m2 is

(a)270

203m/s (b)

271

189m/s (c)

284

195m/s (d) none of above

SECTION IV

This section 2 questions. Each question has four statements ( A, B, C and D) Given in column - I and

five statements ( p,q,r,s and t) in column II. Any given statement in column I can have correct

matching with one or more statement (s) given in column II. For examples, if for a given question,

statement B matches with the statements given in q and r , then for that particular question, against

statement B, darken the bubbles corresponding to q and r in the ORS.

18. A block of mass m is on a rough horizontal floor. P

indicates linear momentum of block in

horizontal direction. Certain cases are given in column-1. Match all the possible types of friction thatcan occur between the block and floor corresponding to the cases.

m Rough floor Column I Column II

(A)dP

dt

constant (p) zero friction

(B)dP

dt

constant > 0 (q) static friction

(C)dP

0dt

(r) kinetic friction

(D) P 0

(s) limiting friction

19. Net force acting on system is f

and its momentum is P

then.Column I Column II

(A) If f

is not zero (p) P

must change magnitude

(B) If f

is zero (q) P

may change direction

(C) If f

is changing in direction only (r) P

may not change direction

(D) If f

is changing in magnitude only (s) P

may not change its magnitude

7/28/2019 mt 3 2010 2

5/27

7/28/2019 mt 3 2010 2

6/27

CHEMISTRY

IIT ians PACE Education Pvt. Ltd. : Andheri / Dadar / Chembur / Thane / delhi / Nerul / Powai/Lucknow

SECTION-II

This section contains 5 questions. The answer to each of the questions is a single digitinteger, ranging from 0 to 9.

26. 4 23 s 3 g 2 g g2NH CO 2NH + CO + H O

Above reaction attains equilibrium at 800K and the value ofKp is 64 atm4. Therefore, the

total pressure of the gases at equilibrium in atm unit is:

27. Before equilibrium is set-up for the chemical reaction 2 4(g) 2 gN O 2NO , vapour density

d of the gaseous mixture was measured. If D is the theoretical value of vapour density,variation of degree of dissociation of N2O4, x with D/d is by the graph. What is value D/d

at point A?

x

D/d

A

28. The ionization energy of a hydrogen-like species is 11757 kJ/mol. If the neutron/protonratio of the species is 1.33, its mass number is______

29. A monoatomic ideal gas undergoes a process in which P2V = constant. The molar heat

capacity of this gas isxR/2. Then value ofx is_______

30. The standard molar enthalpy of formation ofcyclohexane (C6H12)and Benzene (C6H6) at25C are 156 kJ/mole and +49 kJ/mole; respectively. The standard heat of hydrogeneation

of cyclohexene is 119 kJ/mole. Using these data, resonance energy of benzene is foundto be 38x kJ/mole. Calculate the value ofx.

Space for Rough Work

7/28/2019 mt 3 2010 2

7/27

CHEMISTRY


SECTION-III

This section contains 2 groups of questions. Each group has 3 multiple choicequestion based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its

answer, out of which ONLY ONE is correct.


In some situations oxidation state cannot be determined on the basis of simple algebraic

calculation. In such cases, it is formulated on the basis of bonding. For example, peroxoaceticacid, CH3COOOH, the bonding is as follows

HC

CO

OH

H

HO

There exists one per-oxo (OO) linkage, where oxygen is in 1 oxidation state. So the givenacid has one oxygen atom in 2 oxidation state and the other two in 1 oxidation state.

31. Which of the following compounds has a peroxide linkage.(A) SnO2 (B) SrO2

(C) SiO2 (D) SO2

32. The oxidation state of sulphur in Marshalls acid (peroxo disulphuric acid) is

The probable structure of the compound is

HO S

O

O

O O S

O

O

O H

(A) +2 (B) +4(C) +6 (D) +7

33. Select the correct co-efficient in the given reaction, when it is balanced.

x MnO4

+y H2O2 + z H+ p Mn+2 + q O2 + r H2O

(A) x = 1, y = 5, z = 8 (B) x = 5, y = 1, z = 8(C) x = 1, y = 5, z = 6 (D) x = 2, y = 5, z = 6


7/28/2019 mt 3 2010 2

8/27

CHEMISTRY


For dilute solutions active mass is equal to concentration. Taking the example of the reaction.

N2(g) + 3H2(g) 2NH3(g), We can write,Rate of forward reaction rf= kf[N2][H2]

3rate of reverse reaction rr=kb[NH3]

2

When the rate of forward reaction becomes equal to therate of backward reaction, it is known as situation of

equilibrium. Thus at equilibrium

kf[N2][H2]3

=kb[NH3]2

For example in a given reaction, A xB, the

concentration of both A and B and rate of reaction is

measured experimentally with respect to time as given

in graph.

34. Calculate the value of x?

(A) 1 (B)

2 (C)

3 (D) cant be calculated

35. Calculate the value of f bk k for given example,A xB, at given temperature

(A) 5.2 (B) 1.2 (C) 0.5 (D) 2

36. If positive catalyst in introduced in given example to increase the rate of reaction, out of the

following which curve represents the modification with catalyst?

(A) (B)

(C) (D)

SECTION IV

0.1

0.3

0.5

20 40 60 80

B

A

Time (Minutes)

Conc.

(M)

0.1

0.3

0.5

20 40 60 80

B

A

Time (Minutes)

Conc.

(M)

0.1

0.3

0.5

20 40 60 80

B

A

Time (Minutes)

Conc.

(M)

0.1

0.3

0.5

20 40 60 80

B

A

Time (Minutes)

Conc.

(M)

0.1

0.3

0.5

20 40 60 80

B

A

Time (Minutes)

Conc.

(M)

7/28/2019 mt 3 2010 2

9/27

CHEMISTRY


This section 2 questions. Each question has four statements ( A, B, C and D) Given incolumn - I and five statements ( p,q,r,s and t) in column II. Any given statement incolumn I can have correct matching with one or more statement (s) given in columnII. For examples, if for a given question, statement B matches with the statements givenin q and r , then for that particular question, against statement B, darken the bubbles

corresponding to q and r in the ORS.

37. Match the equilibrium mentioned in Column I with their properties mentioned inColumn II

38. Match the reations mentioned in Column I with properties mentioned in Column IICOLUMN - I COLUMN - II

A. P. Redox reaction

B. - -2 3Cl Cl +ClO Q. Nonredox reaction

C. 2O R. Disproportionate reaction

D. 3 4 3 4 22NaCl + Ca PO Na PO + CaCl S. nfactor of reactant = 5/3

T. nfactor of one of reactant =6

COLUMN I COLUMN II

A. 2(g) 2(g) 3(g)N + 3H 2NH ; H = ve P. Low pressure favours the yield of

the product

B. 2(g) 2(g) (g)N + O 2NO ; H = +ve Q. Addition of He at constant pressure

decreases the yield of the product

C. 4 (s) 3 (g) 2 (g)NH HS NH + H S ; H= +ve R. Rate constant of forward reaction

increases more than the increase of the

rate constant of backward reaction with

increase of temperature.

D. 2(g) 2 4 (g)2NO N O ; H = veS. High temperature favours consumption of

product.

T. For the forward reaction:

{Heat of reaction at constant volume} >{Heat of reaction at constant pressure}

7/28/2019 mt 3 2010 2

10/27

PART - I I I MATHS

SECTION - I

T his section contains6 multiple choice questions. Each question has four choices (A ) , (B) , (C ) and

(D ) out of which ONLY ONE is correct.

39. If xlog7

20termsnupto....1074termsnupto.....531

10

and n = .......xlogxlogxlogxlog 81

104

1

102

1

1010 , then x is equal to

(A) 103 (B) 105 (C) 106 (D) 107

40. Inside the unit circle S = {(x, y) | x2 + y2 = 1} there are three smaller circles of equal radius a ,tangent to each other and to S. The value of a equals

(A) 122 (B) 323 (C) 322 (D) 133

41. If the algebraic sum of the perpendicular distances from the points A(0, 2), B(2, 0) and C(1, 1) on a variablestraight line is zero, then the line always passes through a fixed point P, which w.r.t. theABC is its

(A) centroid (B) incentre (C) orthocentre (D) circumcentre

42. As shown in the figure, three circles which have the sameradius r, have centers at (0,0), (1,1)

rr1C

y

2C

Cr

0x

1 2

1and (2,1). If they have common tangent line,

then their radius r is

(A)5 1

2

(B)

5

10

(C)1

2(C)

3 1

2

43. If the circles 0bzazazz and 0dzczczz , b and d are real, cut orthogonally, then caReequal to

(A) b + d (B) b d (C)2

db (D)2

|db|

44. The value of

1n

n1n

5

n)1( equals

(A)12

5(B)

24

5(C)

36

5(D)

16

5

7/28/2019 mt 3 2010 2

11/27

SECTION II

T his section contains a group of5 questions. T he answer to each of the questions is a single - digit

integer, ranging from 0 to 9. T he correct digit below the question no. in the O RS is to be bubbled.

45. If g1, g2, g3 ...........gn are in increasing G.P. such that g1 + gn = 66, g3 gn2 = 128 and 126gn

1ii

.

Find the value of n.

46. If M and m are the maximum and minimum values ofx

yfor pair of real number (x, y) which satisfy the

equation (x 3)2 + (y 3)2 = 6, then find the value of (M + m).

47. If the minimum value of the expression | z |2 + | z 3 |2 + | z 6i |2 is V. Find theV

15

(where z = x + iy, x, y R).

48. The value of the expression 223223log6

cosec9 can be expressed in the lowest form as q

p

where

p and q are relatively prime, then p + q equals.

49. If a, b, c are positive real numbers such that7log3a = 27 ;

11log7b = 49 and25log11c = 11 .

If E =

211

27

23 )25(log)11(log)7(log cba . Find

2E

469

SECTION - III

T his section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choicequestions have to be answered. Each of these questions has four choices (A ) , (B) , (C ) and (D )

out of which ONLY ONE is correct.

Paragraph for question nos. 50 to 52

Let < an> and < bn> be the arithmetic sequences each with common difference 2 such that a1 < b1 and let

n

1k kn

ac ,

n

1k kn

bd . Suppose that the points AAn (an, cn), Bn (bn, dn) are all lying on

C : y = px2 + qx + r where p, q, r are constants.

50. The value of p equals

(A)4

1(B)

3

1(C)

2

1(D) 2

7/28/2019 mt 3 2010 2

12/27

51. The value of q equals

(A)4

1(B)

3

1(C)

2

1(D) 2

52. If r = 0 then the value of a1 and b1 are

(A)2

1and 1 (B) 1 and

2

3(C) 0 and 2 (D)

2

1and 2

Paragraph for question nos. 53 to 55

Consider the family of circles x2 + y2 2x 2y 8 = 0 passing through two fixed points A and B.Also S = 0 is a circle of this family, the tangent to which at A and B intersect on the linex + 2y + 5 = 0.

53. The distance between the points A and B, is

(A) 4 (B) 24 (C) 6 (D) 8

54. The area of an equilateral triangle inscribed in S, is

(A)4

327(B)

2

39(C)

2

327(D) 39

55. If the circle x2 + y2 10x + 2y + c = 0 is orthogonal to S = 0, then the value of c equals(A) 8 (B) 9 (C) 10 (D) 12

SECTION - IV

T his section 2 questions. Each question has four statements ( A ,B, C and D ) G iven in column - I andfive statements ( p,q,r,s and t) in column II . A ny given statement in column I can have correct match-

ing with one or more statement (s) given in column II .For examples, if for a given question, statementB matches with the statements given in q and r , then for that particular question, against statement B ,

darken the bubbles corresponding to q and r in the O RS.

56. Consider the quadratic trinomial f (x) = 2x2 10px + 7p 1, where p is a parameter. Find the range ofpin the following conditions given in column-I.

Column-I Column-II

(A) If both roots of f (x) = 0 are confined in ( 1, 1) then (P)

,

5

2

(B) Exactly one root of f (x) = 0 lies in (1, 1) (Q)

(C) Both roots of f (x) = 0 are greater than 1 (R)

3

1,

17

1

(D) One root of f (x) = 0 is greater than 1 and other root of (S)

,3

1

17

1,

f (x) = 0 is less than 1

(T)

,3

1

17

1,

7/28/2019 mt 3 2010 2

13/27

7/28/2019 mt 3 2010 2

14/27

7/28/2019 mt 3 2010 2

15/27

PAPER -2 (PHYSICS SOLUTIONS)

1 (A) 0 0F T F T

mv v2 2m

Work done = change in kinetic energy =

2

0F T1 m 02 2m

=2 20F T

8m

2 (C)

20 2

yu 20m /s

o45

xu 20m / s

initial velocity

1

u 20i 20 j

a 10 j, t 1s

v u at

v 20i 10 j

v 10 5 m / s

1 1 1 2tan tan , sin cos

2 2 5 5

| v | 10 5

yv 10

ta g sin g

na g cos

x20 v

After 1s

2

t

2

n

22 2

n

10a gsin 2 5 m / s

510 2

a g cos 4 5 m / s5

10 5v va g cos R

10 2R g cos

5

R 25 5 m

7/28/2019 mt 3 2010 2

16/27

3 (B) If massless rod velocity becomes zero at topmost point.Apply work energy theorem

210 m 3gL mgh,

2

1

3 3LmgL mgh, h

2 2

If massless string tension in string becomes zero at topmost point v gLcos .

mgT 0

v

Apply work energy 1 1

mgLcos m 3gL mg L Lcos

2 2

1 8 gLcos , sin , v

3 3 3

2 2

gL 8

v sin 8L 4L3 9

2g 20 54 27

v

L

L

LLcos

3

2 2v sin 4L

2g 27

Max height reached =L 4L

L

3 27

2

40Lh

27

1

2

3Lh 812

40Lh 80

27

7/28/2019 mt 3 2010 2

17/27

4 (C) 21 22

mv vM

Maximum potential energy is when both move with same velocity say u , during the event of

collision. Momentum is conserved.

2

m nm u mm

1 2

1u

n m

Kinetic energy at maximum compression

2

2

1 1 2

2 2 1

m nm u m nm

n m

Maximum potential energy Changes in Kinetic energy

2

1 21

2 11

m n

nn m

1

n

n

5 (C)2

0du b

adx

2 bxa

&a b are off opposition signs. 1, 2a b & 1, 2a b are possible

2

2 3

20

d u b

dx x , 0, 0x b

0b & &a b are of opposite signs

only 1, 2a b is possible

6 (D) Collision between &A B is oblique elastic. Let &x yv v be velocity components of A along horizontal &vertical before collision with B . Collision ofB with &D A with Care elastic head on.

2 2A x yv V V

2 2A D CV V V

7 (4) .dw f d s

.f dx i dy j

xy i dx i dy j

dw xy dx Now2

xy

2

2 2

x x dxdw x dx

;

22 2 3

02 6

x dx xw dw

23

0

1 8 4

6 6 3 3

kx

4k

7/28/2019 mt 3 2010 2

18/27

8 (2) We required power due to friction

f friction cosmg

Instantaneous velocity sin cos g t

Power cos sin cosfv mg g g t

1

1 10 0.8 10 0.6 1 10 0.66

t

2t kt 2k

9 (3)

F

f sin N

1000 f cos

For minimum force, the applied force should be a pull at an angle to horizontal such that tan

2 2

1sin ,cos

1 1

sin 1000N f

2

1000 sin 10001

fN f

Friction2

21000

1

fN

Friction should be equal to cosf

2cos

1

ff

fis 500N

2

2 2

500 5001000

1 1

2

2 2

500 500100

1 1

1

3

2

13

10 (3)1st case 212

mgx kx

0.05x 400k m

2nd case

Now each spring is of length 0

2

L, where 0L is natural length of uncut spring. Spring constant of each

of the 2 parts is 2 800k m . Now in second case,

2 210 1 1mg 0.05 800m 0.05 800m 0.052 2 2

110 0.3m, 0.3 n 10 n 3 .

7/28/2019 mt 3 2010 2

19/27

11 (2)

gsin g

v2

v

R

2

sinv

gR

2

cos mv

mgR

cos sin , 45g g

2

2

R

v

max

1

2

2 2 2

g RR

h R

4 2 4 .

2 4 2 4 2

R R R R RR

5 4 2 5 6 12

6 114 2

R R

2m

12 (A)When A loses contact T = 10 N,F = 3T, F = 30 N

F = 30t 30N = 30t t = 1sWhen B loses contact 2T = 40N

T = 20 NF = 3T = 60N 60 = 30t, t = 2s Time 10g is 2 1 = 1second

13 (A)For getting velocity we have to use calculus, being force is variable.Now acceleration of

dv dv Net force of A 10t 10A a

dt dt mass of A 1

10 t

10

A

1v v t 2s

v 0 t 1

1dv10t 10 dv (10t 10)dt

dt

7/28/2019 mt 3 2010 2

20/27

7/28/2019 mt 3 2010 2

21/27

dP

dt

0 means body is in static or dynamic equilibrium because resultant force on body is zero.

So body may be at rest, may be moving with uniform velocity or may be on verge of moving.

So friction can be anything.

P 0

means bodys velocity is zero. So friction cannot be kinetic.

19 A qrs ;B rs; C qs; D- pqr

(a) If f is not zero P

may not change magnitude as in case of centripetal force in uniform circularmotion For a p is not valid.

(b) If f

is zero, P

will not change magnitude and direction for b r and s are valid.

(c) If f

is changing direction only then it is centripetal force in uniform circular motion, so c q,s .

(d) for f

changing magnitude p, q and r may be valid.

7/28/2019 mt 3 2010 2

22/27

7/28/2019 mt 3 2010 2

23/27

7/28/2019 mt 3 2010 2

24/27

43. (C)

[Sol. The centres of the circles are a and c are radii baa and dcc . The circle cuts orthogonally

| c a |2 = (c a) ac = dccbaa dbacac

caRe =2

db ]

44. (C)

[Sol. We have S =5

1 25

2+ 35

3 45

4+ 55

5+ ............

5

S= 25

1 35

2+ 45

3 55

4+ ............

add

5

S6=

5

1 25

1+ 35

1 45

1+ 55

1 ............

5

S6=

511

5

1

=

6

5

5

1 S =

36

5Ans. ]

45. [Ans. 6][So.. g3 gn2 = 128 = g1 gn (using property of G.P.)

Also g1 + gn = 66 ...........(i)

gn g1 = 1284)66(gg4)gg(2

n12

n1

gng1 = 62 ......(ii) From (i) & (ii), we get

gn = 64, g1 = 2Now, let common ratio of G.P. be r, so

gn = g1 rn1

= 2.rn1

= 64 rn1 = 32 .....(iii)

Now,

n

1iig = 126

)1r(

)1r(2 n

= 126 63

1r

1rn

32 r 1 = 63 (r1) [on using (iii)]

31 r = 62 r = 2 ....(iv)Hence from (iii) & (iv) , we get

2n1 = 32 = 25 n = 6 ]

7/28/2019 mt 3 2010 2

25/27

7/28/2019 mt 3 2010 2

26/27

an = p(an + an 1) (an an 1) + q(an an 1)an = (an an 1) [p(an + an 1) + q] ....(3) (an an 1 = d)

On putting n = 2 and 3 in equation (3), we geta2 = d [p(a2 + a1) + q] .....(4)a3 = d [p(a3 + a2) + q] .....(5)

Now (5) (4), we get

])aa(p[daa

d2

13

d

23

4p = 1 p =

4

1Ans.

(ii) To find q : cn = rqaap n2n

On putting n = 1, 2 in above equation, we get c1 = a1 = raqap 121 ....(1)

c2 = a1 + a2 = raqap 222 ....(2)

but a2 = a1 + 2 2a1 + 2 = p(a1 + 2)2 + q(a1 + 2) + r

= raqap 121 + 4a1p + 4p+ 2q (4p = 1)

2a1 + 2 = c1 + a1 + 1 + 2q ( c1 = a1)

2a1

+ 2 = 2a1

+ 1 + 2q q =2

1 Ans.

(iii) if r = 0, then, c1 = 121

qaap

a1 = 121 a2

1a

4

1 ( c1 = a1)

0a2a 121 a1 = 0 or a1 = 2 Also d 1 = 1

21 qbb4

1

11bdand

2

1q

b1 = 121 b2

1b

4

1

0b2b1

2

1

b1

= 0 or b1

= 2 But a1

< b1

a1

= 0 and b1

= 2 Ans. ]

53. (C) 54. (C)

55. (D)[Sol.(i) We have (x2 + y2 2x 8) 2 y = 0 , which is the form of S + L = 0

On solving S = 0 and L = 0 where S x2 + y2 2x 8 = 0 and L y = 0Put y = 0 x2 2x 8 = 0 (x 4) (x + 2) = 0 x = 2 or 4 A (4, 0) ; B (2, 0) Distance between A and B = 6 Ans.

(ii) Equation of AB which is chord of contact with respect toP(x1, y1) is

xx1 + yy1 1 (x+x1) (y + y1) 8 = 0 x(x1 1) + y (y1 ) (x1 + y1+ 8) = 0 ....(1)P(x ,1 1y )

x+2y+5=0

A(4, 0)B(2, 0)

Also equation of AB is x-axis i.e0x + 1y + 0c = 0 ....(2)

On comparing (1) & (2), we get

0

1x1 =1

y1 =0

8yx 11 = k say

x1 1 = 0 x1 = 1; y1 = + k

7/28/2019 mt 3 2010 2

27/27

and x1 + y1 + 8 = 0 9 + y1 = 0 y1 =

9

Also x1 + 2y1 + 5 = 0 1

18+ 5 = 0 =

6

18 = 3

Required equation of circle S = 0 is x2 + y2 2x 6y 8 = 0

Now radius of circle S = 0 is r = 891 = 23

Hence area of equilateral triangle inscribed in a circle of radius 'r' =2r

4

33= 18

4

33=

2

327Ans.

(iii) If the circle x2 + y2 10x + 2y + c = 0 is orthogonal to S x2 + y2 2x 6y 8 = 0,

so by applying condition of orthogonality, we get )3()1()1()5(2 = c 8 4 = c 8Hence c = 12 Ans. ]

56. [Ans. (A) R; (B) S ; (C) Q ; (D) Q]

57. [Ans. (A) P; (B) R; (C) T; (D) Q, S][Sol.

(A) Let f (x) (x x1)(x x2)(x x3)(x x4)f ( i) = (i + x1)(i + x2)(i + x3)(i + x4)

| f ( i) | = | (i + x1)(i + x2)(i + x3)(i + x4) | = 1x1x1x1x 24232221 = 1 x1 = x2 = x3 = x4 = 0 a = b = c = d = 0

(B) From the figure 3i

e3z

(3,0) (3,0)x

y

O

6030 60120

3

z

arg z = 3

Hence tan2(arg z) 2 cos(arg z) = 21360cos260tan2 (C) By rotation, we get

3

i

ez)1z(

)z()1z(

A(z)

C(z+1)B( z)

3

2z + 1 =

2

3i

2

1 2z =

2

3i

2

1

z =

43i

41 Re(z) =

41 5 + 4

41 = 5 1 = 4 Ans.

(D) We have z1x + z2

x = 2x

( 22)x + ( 2)x = 2x ( 1)x[2x + x] = 1 ( 1)x[2x + x + 1x 1] = 1Clearly x 3n, n I (Because if x is an integral multiple of 3 then LHS RHS)

]11[)1(zero

xxx2x = 1

N if 1 3]

Documents

mt 3 2010 2