Past Year Papers AIEEE 2008

8/10/2019 Past Year Papers AIEEE 2008

1/27

440 AIEEE

AIEEE - 2008 Solution

1 (2) 2 (2) 3 (4) 4 (3) 5 (1) 6 (2) 7 (1) 8 (2) 9 (3) 10 (4)

11 (1) 12 (4) 13 (3) 14 () 15 (1) 16 (3) 17 (4) 18 (1) 19 (3) 20 (3)

21 (3) 22 (3) 23 (3) 24 (2) 25 (1) 26 (4) 27 (4) 28 (2) 29 (1) 30 (1)

31 (3) 32 (3) 33 (2) 34 (4) 35 (1) 36 (4) 37 (4) 38 (3) 39 (2) 40 (1)

41 (4) 42 (2) 43 (1) 44 (4) 45 (2) 46 (1) 47 (2) 48 (3) 49 (4) 50 (3)

51 (2) 52 (2) 53 (3) 54 (3) 55 (3) 56 (2) 57 (1) 58 (4) 59 (3) 60 (1)

61 (3) 62 (4) 63 (2) 64 (4) 65 (1) 66 (3) 67 (2) 68 (4) 69 (1) 70 (3)

71 (2) 72 (2) 73 (3) 74 (1) 75 (2) 76 (3) 77 (4) 78 (3) 79 (4) 80 (4)

81 (4) 82 (4) 83 (4) 84 (4) 85 (3) 86 (1) 87 (2) 88 (4) 89 (2) 90 (4)

91 (1) 92 (3) 93 (2) 94 (1) 95 (2) 96 (2) 97 (1) 98 (3) 99 (2) 100 (4)

101 (3) 102 (4) 103 (4) 104 (3) 105 (4)

PHYSICS

1. For construction interference,

2d cos i =e

hn

2m V = on substituting values we get, V 50V

2. Braggs relation n= 2d sinfor having an intensity maximum for diffraction pattern.

But as the angle of incidence is given, n= 2d cosi is the formula for nding a peak.

3. The electron beam will be diffracted and the maxima is obtained at y = 0. Also the distance

between the rst minima on both side will be greater than d.

4. Mass of planet, Mp= 10 M

e, where M

eis mass of earth.

Escape speed is given by2GM

vR

=

So, for planet vp

p e

p e

2G M 100 2GM

R R

= =

=10 ve= 10 11 km/s = 110 km/s.

5. The forces acting on the solid ball when it is falling through a liquid are mg downwards,

thrust by Archimedes principle upwards and the force due to the force of friction also acting

AIEEE 2008 SOLUTION


2/27

441Past 5 Years Papers: Solved

upwards. The viscous force rapidly increases with velocity, attaining a maximum when the

ball reaches the terminal velocity. Then the acceleration is zero. mg V2g kv2= ma where

V is volume, v is the terminal velocity. When the ball is moving with terminal velocity a = 0.

Therefore, V1g V

2g kv2= 0.

( )1 2Vgvk

=

6. According to the condition of balancing

55 R

20 80=

R = 220

7. CMxdm

Xdm

=

if n = 0, thenCM

LX

2=

As n increases, the centre of mass shifts away fromL

x2

= which only option (1) issatisfying.

8. 1l

RTv

M

= assuming M is the average molar mass of the air (i.e., nitrogen) and is also for

nitrogen.


3/27

442 AIEEE

1 21 2

RT RTv ,v

M M

= = where T

1and T

2stand for winter and summer temperatures.

11

vL 18cm

n 4

= = = . At temperature T

1

At T2, Summer, v

2> v

1.

22

v 3rL 3 18

n 4= = > .

L2> 54 cm.

9. We know that F = q v B

21 1

1

F MLTB MT C

qv C LT

= = =

10. Moment of inertia of square plate about XY is2ma

6. Moment of inertia about ZZ can be

computed using parallel axis theorem

2

ZZ' XY

aI I m

2

= +

2 2 2ma ma 2ma

6 2 3= + =

11. Momentum is mv

m = 3.513 kg : v = 5.00 m/s


4/27


mv = 17.57 Kg m s1

12. The average speed of the athlete

100v 10m / s

10

= =

21K.E. mv2

=

If mass is 40 kg then,

( )21K.E. 40 10 2000 J2

= =

If mass is 100 kg then,

( )21K.E. 100 10 5000 J2

= =

13. 1 2

1 2

C CC

C C=

+ . (i)

where 1 01

K AC

d / 3

= . (ii)

and 2 02

K AC

2d / 3

= . (iii)

It is given that

0A

9pFd

=

On substituting Eqs. (ii) and (iii) in Eq. (i), we get

120A

C 9 10 Fd

= =

With dielectric, 0kA

Cd

=

01

A.3C 9C

d / 3

= =


5/27

444 AIEEE

02

A.6C 9C

2d / 3

= =

1 2

total1 2

C C

C C C = + as they are in series.

9C 9C 9C

18C 2

= =

129 9 10 F2

totalC 40.5pF. =

14. No option is correct

RTvM

=

1 1 2

2 2 1

74

v M 55v M

323

= =

2

460 21

v 25 8=

2460 5 2 2

v 142021

= =

15. We know that energy is released when heavy nuclei undergo ssion or light nuclei undergo

fusion. Therefore statement (I) is correct.

The second statement is false because for heavy nuclei the binding energy per nucleon

decreases with increasing Z and for light nuclei, B.E/nucleon increases with increasing Z.

16. Correct option is (3) you can make an analogy with Gausss law in electrostatics.

17. The liquid 1 is over liquid 2. Therefore, 1<

2. If

3had been greater than

2, it will not be

partially inside but anywhere inside liquid 2. If 3=

2or it would have sunk totally if

3had

been greater than 2

.

1<

3<

2.


6/27


18. It is p-n-p transistor with R as base.

19. Electric eld at a distance r from A is,2

1E

2 r=

dV Edr = a b

C B 2

a

I drV V V

2 r

+ =

I 1 1V

2 a a b

= +

20. From2

IE J

2 r

= =

21. According to the new Cartesian system used in schools,1 1 1

v u f = for a convex lens. u has to

be negative.If v = , u = f and if u = , v = f

A parallel beam (u = ) is focused at f and if the object is at f, the rays are parallel. The point

which meets curve at u = v gives 2f.

Therefore, v is +ve, u is negative, both are symmetrical and this curve satises all the

conditions

for a convex lens.

22. Initial kinetic energy of the system

( )2211 1

K.E mu M 02 2

= +

1 0.5 2 2 0 1J2

= + =

For collision, applying conservation of linear momentum m u = (m+M) v0.5 2 = (0.5 + 1) v

2v m / s

3 =

Final kinetic energy of the system is

( ) 2f1

K.E m M v2

= +


7/27

446 AIEEE

( )1 2 2 10.5 1 J2 3 3 3

= + =

Energy loss during collision1

1 J

3

=

= 0.67 J

23. Soap solution has lower surface tension, T as compared to pure water and capillary rise

2Tcosh

rg

=

so h is less for soap solution.

24. Supposing that the force of attraction in Bohr atom does not follow inverse square law but

inversely proportional to r,

2

0

1 e

4 rwould have been

2mv

r=

22

0

emv k

4 = =

21 1mv k2 2

=

This is independent of n.

From mvrn=

nh

2

as mv is independent of n, rnn.

25. y(x,t) = 0.005 cos (x t) (Given)Comparing it with the standard equation of wave y(x,t) = a cos (kx t) we get

k = and =

But2

k

=

and2

T

=

2 =

and

2

T

=

Given that= 0.08 m and T = 2.0 S

2 250.08

= = and 22 = =

26. The mutual inductance 0 1 2N N A

Ml

=

where N1= 300 turns, N

2= 400 turns, A = 10 cm2and l = 20 cm.

Substituting the values in the given formula, we get M = 2.4 104H.

27. It is OR gate. When either of them conducts, the gate conducts.

28. For the body starting from rest


8/27


2

1

1x 0 at

2= +

2

1

1x at

2

=

For the body moving with constant speed

x2= vt

2

1 2

1x x at vt

2 =

at t = 0, x1x

2= 0

Forv

ta

< ; the slope is negative

Forv

ta

= ; the slope is zero

Forv

ta

> ; the slope is positive

These characteristics are represented by graph (2).

29. Conceptual.

30. The electric eld for a uniformly charged spherical shell is given in the gure. Inside the shell,

the eld is zero and it is maximum at the surface and then decreases 1/r2.

2

0

QE

4 .r=

outside shell and zero inside.

31. Applying Kirchhoffs loop law in AB P2P

1A we get


9/27

448 AIEEE

2i + 5 10 i1= 0 .(i)

Again applying Kirchhoffs loop law in P2CDP

1P

2we get

10 i1+ 2 i + i

1= 0 .(ii)

From (i) and (ii)

11

5 10i11i 2 0

2

+ =

1

1i 0.03

32 = = A from P

2to P

1.

32. 0i

B2 R

=

.

Direction is given by Right hand palm value No. 1

762 10 100B T 5 10 T

4

= = towards south.

33. The value of relative permeability of diamagnetic materials are slightly less than 1 and ris

quite high. According to the values given, we take r= 1.5 and

r= 0.5.

34. Least count of screw gauge

0.5mm 0.01mm

50= =

Reading = [Main scale reading + circular scale reading L.C] (zero error)

= [3 + 35 0.01] (0.03) = 3.38 mm

35. As no work is done and system is thermally insulated from surrounding, it means sum of

internal energy of gas in two partitions is constant ie, U = U1+ U

2


10/27


Assuming both gases have same degree of freedom, then

( )1 2f n n RTU

2

+= and 1 1

1

fnRTU

2=

2 22 fn RTU

2=

Solving we get, 1 2 1 1 2 2

1 1 1 2 2 2

T T (P V P V )T

P V T P v T

+=

+

CHEMISTRY

36. SN2 attack occurs if the backside route of attack is not sterically hindered by substituents

on the substrate. Therefore this mechanism usually occurs at an unhindered primary carbon

centre. Lesser the steric hindrance better the SN2 reaction. So ease of reaction is 1> 2>

3SN2 involves inversion of conguration stereo chemically.

(Remember ! molecule is optically active, optical inversion, Walden inversion is known)

37. In the given complex [E(en)2(C

2O

4)]+NO

2ethylene diamine is a bidentate ligand and (C

2O

42)

oxalate ion is also bidentate ligand. Therefore, co-ordination number of the complex is 6 i.e., it is

an octahedral complex. Oxidation number of E in the given complex is x + 2 0 + 1 (2) = +1

x = 3

38. Ozone layer acts as a shield and does not allow ultraviolet radiation from sun to reach earth.

It does not prevent infra-red radiation from sun to reach earth, thus option (3) is wrong

statement and so it is the correct answer.

39.

First sulphonation is the means to block para position and to reduce the reactivity of phenolic

ring against strong oxidizing agent HNO3. (The use of conc HNO

3 over phenol cause the

oxidation of ring mainly). The strong acidic medium in second step cause desulphonation

(ipso mechanism) also.

40. The reaction sequence is as follows :


11/27

450 AIEEE

41. Completing the sequence of given reaction,

3O

3 3CH CH CH CH =

3 22CH CHO H O ZnO

'B '

+ +

Thus B is CH3CHO.

42. Lanthanides : 1 14 0 1 2[Xe]4f 5d 6s

Actinides : 1 14 0 1 2[Rn]5f 6d 7s

Lanthanides and actinides use core d and f orbitals also to show higher oxidation state. As

actinides have comparatively low energy difference between f and d orbitals show more

oxidation states.

43. CFSE (crystal eld splitting energy) for octahedral complex, 0 depends on the strength

of negative ligand. Spectrochemically it has been found that the strength of splitting is as

follows:

2 3

2 2

2

2

CO CH NO en NH py

NCS H O O OX OH F

Cl SCN S Br I

> > > > >

> > > > > >

> > > >

44. We have,0 0

A B

P 520mmHgandP 1000mmHg= =


12/27


Let mole fraction of A is solution XA

and mole fraction of B is solution XB

Then, at 1 atm pressure i.e. at 760 mm Hg

0 0

A A B BP X P X 760mmHg+ =

( )0 0A A B AP X P 1 X 760mmHg + =

A A520X 1000 1000X 760mmHg + =

A

1X

2 =

or 50 mole per cent.

45.1

A 2B2

Remember for aA bB

1 d[A] 1 d[B]

a dt b dt = = Rate of reaction

For the given reaction

2d[A] 1 d[B]

dt 2 dt = = Rate of reaction

Rate of disappearance of A

d[A] 1 d[B] 1 d[B]

dt 2 2 dt 4 dt= = =

46.X 2Y Z P Q

Initialmol. 1 0 1 0 0

At equilibrium 1 2 1

+

1

2

2 1

YP

X1

2P

P 1K

1PP

1

+ = =

+

and

2

2 2P Q

P

Z2

P PP P 1 1

K1P

P1

+ + = =

+

1

2

1P 2

4 PK

1

=

. (i)

2

2

2P 2

PK

1

=

.(ii)


13/27

452 AIEEE

1

2

P

P

K 1

K 9=

.(iii)

Substituting values of1P

K and2P

K from equation (i) and (ii) into (iii), we get2

12

2

2

2

4 P119P

1

=

1

2

4P 1

P 9 = 1

2

P 1

P 36 =

47. The energy involved in the conversion of

12

1Cl (g) toCl (aq)

2 is given by

( ) ( ) ( )2diss Cl EA Cl hyd Cl

1H H H H

2

= + +

Substituting various values from given data, we get

( ) ( )11H 240 349 381 kJmol

2 = + +

= (120 349 381) kJ mol1

= 610 kJ mol1

48. 2MS + C 2 M + CS2 G

1= positive

2MO + C 2 M + CO2 G

2= negative

The value of G for the formation of CO2is negative, ie, it is thermodynamically more stable

than CS2. Also metal sulphides are thermodynamically more stable than CS

2. Metal sulphides

are more stable than the corresponding oxides, so they are roasted to convert into less stable

oxides.49. Bakelite is a thermosetting polymer which is made by reaction between phenol and

formaldehyde.

50. As equation III is obtained on adding equation I and equation II, so K3= K

1.K

2.

51.

52.

This structure will be of lowest energy due to resonance stabilization of +ve charge. In all

other three structures, the presence of electron withdrawing NO2group will destabilize the

+ve charge and hence they will have greater energy.

53. Since- D (+) glucose and D (+) glucose differ in conguration at C 1 atom so they are anomers.

Note :- Anomers are those diastereomers that differ in conguration at C 1 atom.


14/27


54.2 2 3

1 3X Y XY

2 2+

reaction products reac tantsS S S =

1reaction

3 1S 50 40 60 40 Jmol

2 2 = + =

G = H TS

At equilibrium as G = 0

H = TS

3H 30 10T 750K

S 40

= = =

55. HSO3F is the super acid. Its acidic strength is greater than any given species. The pK

avalue

of other species are given below :

3

3

4

HCO 10.25

H O 1.74

HSO 1.92

+

Lesser the pKavalue, higher will be its acidic strength. Hence sequence of acidic strength will

be3 3 4 3

HSO F H O HSO HCO+ > > >

56. Isoelectronic species have same number of electrons,

NO+, 22

C , CNand N2all have 14 electrons.

57. CNand NO+both have same number of electrons and same bond order (3).

58. Ionisation enthalpy of hydrogen atom is 1.312 106J mol. It suggests that the energy of

electron in the ground state (rst orbit) is 1.312 106J mol1

E = E2 E

1

6 6

2

1.312 10 1.312 10

12

=

= 9.84 105J mol1

59. The correct formula of inorganic benzene is B3N

3H

6so (4) is incorrect statement.

Boric aci3 3

OH

|

(H BO orB OH)

|

OH

is a lewis acid so (1) is incorrect statement.

The coordination number exhibited by beryllium is 4 and not 6 so statement (2) is incorrect.Both BeCl

2and AlCl

3exhibit bridged structures in solid state so (3) is correct statement.

60. From the given representation of the cell, Ecell

can be found as follows


15/27

454 AIEEE

2 3

23

0 0

cell 3Fe /Fe Cr /Cr 2

Cr0.059E E E log

6 Fe+ +

+

+

=

[Nernst Equ.]

( ) ( )( )

2

3

0.10.0590.42 0.72 log

6 0.01=

=0.059 0.1 0.1

0.42 0.72 log6 0.01 0.01 0.01

+

2

6

0.059 10 0.0590.3 log 0.3 4

6 610

= =

= 0.30 0.0393 = 0.26 V

61. Titration of oxalic acid by KMnO4in the presence of HCl gives unsatisfactory result because

HCl is a better reducing agent than oxalic acid and HCl reduces preferably 4MnO to Mn2+.

62. Moles of glucose18

0.1180

= =

Moles of H2O

178.29.9

18= =

According to Raoults law

0s

solute0

P PX

P

=

s17.5 P 0.1

17.5 10

=

So, Ps= 17.325 mm Hg

63. The cross linked polymers will be formed by RSiCl3


16/27


64. Carbon monoxide is oxidized to carbon dioxide by passing the gases and steam over an iron

oxide or cobalt oxide or chromium oxide catalyst at 673 K resulting in the production of more

H2.

2 3Fe O ,CoO

2 2 2673KCO H O CO H+ +

NaOH

2 2 3alkaliCO Na CO

CO2is absorbed in alkali (NaOH).

The entire reaction is called water gas shift reaction.

65. Number of Y atoms per unit cell in ccp lattice (N) = 4

number of tetrahedral voids = 2N = 2 4 = 8

number of tetrahedral voids occupied by X = 2/3rdof the tetrahedral void =2 16

83 3

= Hence,

the formula of the compound will be

16/3 4 4 3X Y X Y=

66. Higher the gold number, lesser will be the protective power of colloid.

So, A > > > =

70. For salt of weak acid and weak base

a w b

1pH [logK logK logK ]

2= +

a w b

1 1 1pK pK pK

2 2 2= +

1 14.80 7 4.78

2 2= +

= 7.01.

MATHEMATICS

71. In ABC, BC = h cot 60

and in ABD, BD = h cot 45

Since, BD BC = DC


17/27

456 AIEEE

h cot 45 h cot 60= 7

7 7

hcot 45 cot 60 1

13

= =

( )7 3 3 1 7 3 3 1 m23 1 3 1+= = + +72. From the denition of independence of events

P(A B)P(A /B)

P(B)

=

Then P(B). P(A/B) = P(AB) .(i)

Interchanging the role of A and B in (i)

P(A)P(B/A) = P(B A) ..(ii)

As A B = B A, we have from (i) and (ii)P(A)P(B/A) = P(B)P(A/B)

1 2 1. P(B).

4 3 2 =

1 2 1P(B) . .2

4 3 3 = =

73. A number is greater than 3

3 1P(A)

6 2 = =

B number is less than 5

4 2P(B)

6 3 = =

A B number is greater than 3 but less than 5.

1P(A B)

6 =

P(A B) = p(A) + p(B) p(A B)

1 2 1 3 4 11

2 3 6 6

+ = + = =


18/27


74. Since,a

ae 4e

= and1

e2

=

a2a 4

2 =

3a42 =

8a

3 =

75. The vertex is the mid point of FN, that is, vertex = (1, 0)

76. Given equation can be rewritten as

(x + 1)2+ (y + 2)2= ( )22 2 Let required point be Q (, )

Then, mid point of P(1, 0) and Q(, ) is the centre (-1, -2) of the circle

11

2

+ = and

02

2

+=

= 3 and = 4

Required point is (3, 4)

77. Since, Y = {y N : y = 4x + 3 for some x N}

Y = {7, 8, 11, 15, ., }

Let y = 4x + 3

y 3x

4

=

Inverse of f(x) is

y 3

g(y) 4

=


19/27


20/27


21/27

460 AIEEE

Point17 13

0, ,2 2

satises Eq. (i), we get

17 13b 13 2 2

2 1 b a 1

= =

15

2a 1 5

3

2

= =

a = 6

Also,3(1b) =17

2 b2

3b 3 = 17 2b 5b = 20

b = 4

86. Given,x 1 y 2 z 3

k 2 3

= = .(i)

andx 2 y 3 z 1

3 k 2

= = .(ii)

Since, lines intersect at a point

k 2 3

3 k 2 0

1 1 2

=

( ) ( ) ( )k 2k 2 2 6 2 3 3 k 0 + =

22k 5k 25 0 + =

22k 5k 25 0 + =

22k 10k 5k 25 0 + =

2k(k 5) 5(k 5) 0 + + =

5k , 5

2 =

Hence, integer value of k is 5.

87. Statement 2 is ( )n n 1 n 1,n 2+ < +

n n 1,n 2 < + which is true.

2 3 4 5 ............ n < < < <


22/27


Now, 2 n

3 n

1 1n n

n n

Also,1 1

1 n>

adding all, we get

1 1 1 1 n....... n

1 2 3 n n+ + + + > =

Hence, both the statements are correct and statement 2 is a correct explanation of statements

-1.

88. Leta b

Ac d

=

2a b 1 0

c d 0 1

=

(Since, A2= I)

2

2

a bc ab bd 1 0

0 1ac cd bc d

+ + =

+ +

b(a + d) = 0, c (a + d) = 0

and a2+ bc = 1, bc + d2= 1

a = 1, d = 1, b = c = 0

If1 0

A0 1

=

, then

2 1 0 1 0 1 0A I0 1 0 1 0 1

= = =

A I, A I

det A = 1 (statement I is true)

Statement II tr (A) = 1 1 = 0, Statement II is false.

89. We have

( )n n n

n r n r n r

r r rr 0 r 0 r 0

r 1 C x r. C x C x= = =

+ = +

( )n

nn 1 r

r 1r 1

nr. C x 1 x

r

=

= + +


23/27

462 AIEEE

( )n

nn 1 r 1

r 1r 1

nx C x 1 x =

= + +

= nx (1 + x)n1+ (1 + x)n= RHS

statement 2 is correct

Putting x = 1, we get

( ) ( )n

n n 1 n n 1

rr 0

r 1 C n.2 2 n 2 .2

=

+ = + = +

Statement I is also true and statement 2 is a correct explanation for statement I.

90. The given statement r p q

The statement 1 is r1(p q) (p q)

The statement 2 is r2(p q) = (p q) (p q)

we can establish that r = r1

Thus statement 1 is true but statement 2 is false.

91. Since, the number of ways that child can buy the six ice-creams is equal to the number of

different ways of arranging 6 As and 4 Bs in a row.

Number of ways to arrange 6As and 4Bs in a

row 104

10!C

6!4!= =

and number of integral solutions of the equation 1 2 3 4 5x x x x x 6+ + + + = is given by

6 5 1 10 10

5 1 4 5C C C+

=

Statement I is false and statement II is true.

92. We have f (x) =

1(x 1)sin ,if x 1

x 1

0 , if x 1

=

h

f(1 h) f(1)Rf '(1) lim

h+

=

h h

1

hsin 0 1hlim limsinh h

= = ,

which does not exist.

f is not differentiable at x = 1

Also,( ) ( ) x 0

1 x 1 1f '(0) sin cos

x 1 x 1 x 1 =

=

= sin 1 + cos 1

f is differentiable at x = 0


24/27


93. Let the GP be a, ar, ar2, ar3, .

We have a + ar = 12 ....(1)

r2+ ar3= 48 .(2)

on division we have

( )( )

2ar 1 r 48

a 1 r 12

+=

+

r2= 4

r = 2

But the terms are alternately positive and negative,

r = 2

Now12 12

a 121 r 1 2

= = = +

[From (1)]

94. Denote x3 px + q by f(x)

i.e., f(x) = x3 px + q

Now for expression, f(x) = 0, i.e. 3x2 p = 0

p px ,

3 3=

f(x) = 6x

pf " 03


25/27

464 AIEEE

97. 1 15 2

cot cosec tan3 3

+

1 13 2

cot tan tan4 3

= +

1 1

3 2174 3cot tan cot tan

3 2 61

4 3

+

= =

1 6 6cot cot17 17

= =

98. The equation to circle is

(x )2+ (y 2)2= 25 .(1)

Differentiating w.r.t.x

( ) ( )dyx y 2 0

dx + =

( )dyx y 2dx

= .(2)

From (1) and (2) on eliminating

( ) ( )

22 2dy

y 2 y 2 25dx

+ =

(y 2)2(y)2= 25 (y 2)2

99. Since,1 1

0 0

sinx xI dx dx

x x= sin x.

1 13/2

0 0

2I x dx x

3 < =

2I

3 <

and1

1 12

0 0

cosxJ dx x dx 2

x

= < =

J < 2.

100. We have the two curve equations x + 2y2= 0 and x + 3y2= 1

Solving the two equations we get the points of intersection as (2, 1), (2, 1)


26/27


The required area is AOBDA, given by

( )131

2 2

1 1

5y1 3y 2y dy y

3

= =

5 5 2 41 1 2 sq.units

3 3 3 3

= + = =

101. Letsinx

I 2 dx

sin x4

=

Put x t4 =

dx = dt

sin t dt4

I 2sint

+

=

1 12 cot t dt

2 2

= +

= t + log |sin t| + c

x log sin x c4

= + +

102. Leaving S, we have 7 letters M, I, I, I, P, P, I

Number of ways of arranging them =7

7.5.32 4

=

And four S can be put in 8 places in 84

C ways.

The required number of ways = 7.5.3. 84

C

= 7. 64

C . 84

C


27/27

466 AIEEE

103. Given equations are x cy bz = 0

cx y + az = 0

and bx + ay z = 0

For non-zero solution

1 c b

c 1 a 0

b a 1

=

1(1 a2) + c(cab) b(ac + b) = 0

1 a2 c2 abc abc b2= 0

a2+ b2+ c2+ 2abc = 1

104. All entries of square matrix A are integers, therefore all cofactors should also be integers.

If det A = 1 then A1exists. Also all its entries of A1are integers.

105. Let the roots of equation x2 6x + a = 0 be and 4and that of the equation x2 cx + 6 =0 be and 3.

Then + 4= 6

4= a

and + 3= c : 3= 6

a = 8

The equation becomes x2 6x + 8 = 0

(x 2) (x 4) = 0

Roots are 2 and 4

= 2, = 1

common root is 2.

Date post:	02-Jun-2018
Category:	Documents
Upload:	shaharukh-nadaf
View:	217 times
Download:	0 times

Past Year Papers AIEEE 2008

Documents