Past Year Papers AIEEE 2009

8/10/2019 Past Year Papers AIEEE 2009

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396 AIEEE

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

11 (4) 12 (1) 13 (3) 14 (3) 15 (3) 16 (2) 17 (1) 18 (1) 19 (2) 20 (1)

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

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

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

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

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

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

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

1.21

h gt ,(parabolic)2=

v = -gt and after the collision, v = gt (straight line)

Collision is perfectly elastic then ball reaches to same height again and again with same

velocity

2.( )2

GMg' ,

R h=

+acceleration due to gravity at height h

( )

22

2 2

g GM R R. g

9 R hR R h

= = + +

21 R

9 R h

=

+

R 1

R h 3 =

+

3R = R + h

2R = h

3.

AIEEE 2009 SOLUTION


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397Past 5 Years Papers: Solved

TQ KA

X

=

Q constant

X

constant.

4. W = QdV = Q(Vq V

p)

= - 100 (1.6 10-19) (- 4 10)

= + 100 1.6 10-19 14

= +2.24 10-16J

5. Net magnetic eld due to loop ABCD at O is

B = BAB

+ BBC

+ BCD

+ BOA

= 0 0I I

0 04 a 6 4 b 6

+ +

( )0 0 0I I I b a24a 24b 24ab

= =

6. The forces AD and BC are zero because magnetic eld due to a straight wire on AD and BC is

parallel to elementary length of the loop.

7. WAB

= Q - U = nCpdT nC

vdt(at constant pressure)

= n(Cp C

v) dt

= nR dt = 2 R (500 300)

= 400 R


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4/23


Qq

2 2 = or

Q2 2

q=

12. Thermal energy corresponds to internal energy

Mass = 1 kg

Density = 4 kg m-3

3mass 1

Volume mdensity 4

= =

Pressure = 8 104N m-2

45

Internalenergy P V 5 10 J2

= =

13.1

1

E 12l 6 A

R 2= = =

22 2

dlE L R l

dt= +

( )ct/ t2 0l I 1 e=

0

E 12I 6A

R 2 = = =

3

c

L 400 10t 0.2

R 2

= = =

( )t/0.2

2I 6 1 e= Potential drop across

L = E R2I

2= 12 2 6(1 e-5t)=12e-5t

14. (The relation 0R R (1 t)= + is valid for small values of t and also ( )0R R should bemuch smaller than R

0. So, statement (1) is wrong but statement (2) is correct.

15. IR corresponds to least value of2 2

1 2

1 1

n n

i.e., from Paschen, Bracket and P fund series.


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400 AIEEE

Thus the transition corresponds to 5 3.

16. 1 23 4 =

2 13 3

5904 4

= =

1770

442.5nm4

= =

17. u 3i 4j;a 0.4i 0.3j= + = +

v u at= +

( )3i 4 j 0.4i 0.3j 10= + + + 3i j 4i 3j 7i 7j= + + + = +

2 2Speed is 7 7 7 2unit+ =

18. 20

1mv eV 1.68eV

2= =

hc 1240evnm

hv 3.1eV400nm

= = =

03.1eV W 1.6eV = +

0W 1.42eV=

19. Maximum number of beats = v + 1 (v 1) = 2

20. Motor cycle, u = 0, a = 2 ms-2

observe is in motion and source is at rest

0

s

v vn' n

v v

=

+

0330 v94

n n100 330

=

0

330 94330 v

100

=

10

94 33 33 6v 330 ms

10 10

= =

2 2v u 9 33 33 9 1089

s 98m2a 100 100

= = =

21. 1st reaction is fusion and 4threaction is ssion.

22.3

sinC2

=


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sin r = sin (90 - C) =1

cosC2

=

2

1

sinsinr

=

2 1

sin23

=

11

sin3

=

23.1 1 2 2

A l A l=

1 1 1 12

1

A l A l llA 3A 3

= = =

1

2

l3

l =

11 1

Fx l ....(i)

A =

22 2

Fx l ....(ii)

3A =

Here,1 2

x x =

2 12 1

F Fl l

3A A=

12 1 1

2

lF 3F 3F 3 9F

l= = =

24. Work done by conservative force does not depend on the path. Electrostatic force is a

conservative force.

25. Truth table

26.2 2 2

2

aT xT 4 4T Cons tan t

x x TT

= = = =


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402 AIEEE

27. T.E.i= T.E.

r

21

l mgh2

=

2 21 1

ml mgh2 3 =

2 21 l

h6 g

=

28. It is possible when object kept at centre of curvature

u = v

u = 2f, v = 2f

29. Given gure is half wave rectier.

30.

1r 2

402

1

0

Qr4 r dr

RE4 r

=

2

1

4

0

QrE

4 R =


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CHEMISTRY

31.

32. XeF6has much tendency to hydrolyse. The reverse reaction is more spontaneous.

XeF6+ 3H

20 XeO

3+ 6HF

33.

34. Adsorption is an exothermic process i.e, energy is released against van der Waals force ofattraction (physisorpions). Hence, H is always negative.

35. Optical isomerism is usually exhibited by octahedral compounds of the type [M(AA)2B

2], where

(AA) is a symmetrical bidentate ligand. Square plane complexes rarely show optical isomerism

on account of presence of axis of symmetry. Thus among the given options, [Co(en)2(NH

3)

2]3+

exhibits optical isomerism.

36. 2 2sp 3 3

K BaCO Ba CO+ =


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404 AIEEE

9

sp2

423

K 5.1 10Ba

1 10CO

+

= =

[Ba2+] = 5.110-5M

37.34

27 3

h 6.63 10

mv 1.67 10 1 10

= =

= 3.97 10-10meter = 0.397 nanometer.

38. Lower oxidation state of an element forms more basic oxide and hydroxide, while the higher

oxidation state will form more acidic oxide/hydroxide. For example,

39. Given, velocity of e-, v = 600 ms-1

Accuracy of velocity = 0.005%

600 0.005v 0.03

100

= =

According to Heisenbergs uncertainty principle,

hx.m v

4

.

34

31

6.6 10

x 4 3.14 9.1 10 0.03

=

= 1.92 10-3m.

40. Linkage isomers are caused due to presence of ambidentate ligands [Pd(PPh3)

2(NCS)

2] and

[Pd(PPh3)

2(SCN)

2] are linkage isomers due to SCN, ambidentate ligands.

41. In BF3there is signicant pp interaction between unshared p-orbital (having no electron)

over boron and the lone pair of electron over uorine in 2p-orbital.

42.No.ofbondingelectrons No.of antibondingelectorns

Bondorder2

=

Bond order in2

10 7O 1.5

2

+ = =


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Bond order in2

10 7O 1.5

2

= =

Bond order in 22

10 8O 1

2

= =

Bond order in 22

10 4O 3

2

+ = =

Since, bond order1

Bondlength

Bond length is shortest is 22

O +

43. Since the compound formed has a fruity smell, it is an ester, thus the liquid to which ethanoland conc. H

2SO

4are added must be an acid.

2 4concH SO3 2 5

3 2 5 2

CH COOH C H OH

CH COOC H H O

+

+

44.

3 2 3 2CH CH Cl CH CH OH

2 2 2 2

ClCH CH Cl HOCH CH OH

3 2 3 2

intermediate

3acetaldehyde

CH CHCl CH CH(OH)

CH CH O

=

45. BunaN is a copolymer of butadiene (CH2= CH CH = CH

2) and acrylonitrile (CH

2= CHCN).

46. Glucose is considered as a typical carbohydrate which contains CHO and OH group.

47. In group 15 hydrides, the basic character decreases on going down the group due to decrease

in the availability of the lone pair of electrons because of the increase in size of elements from

N to Bi. Thus, correct order of basicity is NH3> PH

3> AsH

3> SbH

3.

48. n-heptane and ethanol form non-ideal solution, as resultant n-heptane-ethanol molecular

interaction is very poor in comparison to ethanol-ethanol or n-heptane so gives positivedeviation.

49. Ionic radii increases on descending the group because of increase in the number of shells;

however it decreases across the period due to increase in effective nuclear charge because

electrons are added in the same shell. However, remember that Li+and Mg2+are diagonally

related.

50. The groups having +I effect decrease the stability while groups having I effect increase the

stability of carbanions. Benzyl carbanion is stabilized, 2 carbanions are more stable, thus the

decreasing order of stability is

( ) ( )3 6 5 2 3 32 3CCl C H CH CH CH CH C> > >


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406 AIEEE

51. Most of the Ln3+ compounds except La3+ an Lu3+ are coloured due to the presence of

f-electrons.

52. When two groups attached to a double bonded carbon atom are same, the compound does

not exhibit geometrical isomerism. Compounds in which the two groups attached to a double

bonded carbon are different, exhibit geometrical isomerism, thus only 2-butene exhibits cis-trans isomerism.

53.

exhibits both geometrical as well as optical isomerism.cis R cis Strans R trans S

54. In cannizaro reaction the transfer of H-to another carbonyl group is difcult and slowest step.

55. The reaction for the formation of(aq)

OH is

( ) ( )2(g) 2(g) aq aq1

H O H OH2

+ + +

This is obtained by adding the two given equations

H for the above reaction = 57.32 + (286.2)

= 228.88 kJ.

56. For fcc unit cell, 4r 2a=

2 361r 127pm

4

= =


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57. % efciency of the fuel cellG

100

=

The concerned reaction is

( )3 2 2 23CH OH l O (g) CO (g) 2H O(l)2

+ +

( ) ( )r f 2 f 2G G CO ,g 2 G H O,l = +

( ) ( )f 3 f 23

G CH COOH,l G O ,g2

= 394.4 + 2(237.2)(166.2)0=702.6

Thus, % efciency is 97

58. 0 0T x X Y yP p x p x= +

where, PT= Total pressure

0

Xp = Vapour pressure of X in pure state.

0

Yp = Vapour pressure of Y in pure state.

xX= Mole fraction of

1X

4=

xY= Mole fraction of

3Y

4=

(i) When T = 300 K, PT= 550 mm Hg

0 0

X Y

1 3550 p p

4 4

= +

0 0

X Yp 3p 2200 ....(1) + =

(ii) When at T = 330 K, 1 mole of Y is added,

( )TP 550 10 mmHg= +

X Y1 4

x andx5 5

= =

0 0X Y1 4560 p p5 5

= +

0 0

X Yp 4p 2800 ....(2) + =

On solving equations (1) and (2), we get

0

Yp 600mmHg= and 0

Xp 400mmHg=

59. Given,

Fe3++ 3eFe E1 = 0.036 V .(i)

Fe2++ 2eFe E2 = 0.439 V .(ii)


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408 AIEEE

We need to calculate

Fe3++ eFe2++ E3 = ? .(iii)

We can obtain (iii) by subtracting (ii) from (i) but E3, we can not obtain that way because

electrode potential is intensive property. Thats when we determine E3 calculating G

3= G

1

G2(G is an extensive property)G

3= 3 0.036 F 20.439 F

G3= 0.108 F 0.878 F

1 F E3 = 0.770 F

E3 = 0.770 F.

60. For rst order reaction,

2.303 100k log

t 100 99=

0.693 2.303 100

log6.93 t 1=

0.693 2.303 2

6.93 t

=

t = 46.06 min

MATHEMATICS

61. The truth table for the logical statements, involved in statement 1, is as follows :

p q q p q (p q) p q

T T F F T TT F T T F F

F T F T F F

F F T F T T

We observe the columns for (p q) and p q are identical, therefore, (p q) isequivalent to (p q)

But (p q) is not a tautology as all entries in its column are not T.

statement 1 is true but statement 2 is false.

62. |adj A| = |A|n-1= |A|2-1= |A|

adj (adj A) = |A|n2A = |A| A = A

63. The solution of f(x) = f1(x) are given by

f(x) = x, which gives (x + 1)2 1 = x

(x + 1)2 (x + 1) = 0

(x + 1)x = 0

x = 1, 0

But as no co-domain of f is specied, nothing can be said about f being ONTO or not.

64. For the numbers 2, 4, 6, 8,.., 2n


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( )( )

2 n n 1x n 1

2n

+ = = +

And ( ) ( )2

2 2x x xVar x2n n

= =

( ) ( )( ) ( )2

2 24n n 1 2n 14 nn 1 n 1

n 6n

+ += + = +

( )( ) ( )22 2n 1 n 1

n 13

+ += +

( ) 4n 2 3n 3n 13

+ = +

( )( ) 2n 1 n 1 n 1

3 3+ = =

statement 1 is false.Clearly , statement 2 is true.

65. f(x) = x|x| and g(x) = sin x

( ) 2

2

sinx , x 0gof(x) sin x | x |

sinx , x 0


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410 AIEEE

( )1y 3 x 2

2 =

x 2y + 4 = 0

The area of the bounded region

( ) ( )3

2

0

y 2 1 2y 4 dy = +

( )3

2

0

y 6y 9 dy= +

( )3

2

0

y 3 dy=

(Let 3 y = t)

( )3

2

0

3 y dy=

33 3 32

0 0

t 3t dt 9

3 3

= = = =

67. We have P(x) = x4+ ax3+ bx2+ cx +d

P(x) = 4x3+ 3ax2+ 2bx + c

But P(0) = 0c = 0

P(x) = x4+ ax3+ bx2+ d

As given that P( 1) < P (1)

1 a + b + d < 1 + a + b + da > 0

Now P(x) = 4x3+ 3ax2+ 2bx= x(4x2+ 3ax + 2b)

As P (x) = 0, there is only one solution x = 0, therefore, 4x2+ 3ax + 2b = 0 should not have

any real roots i.e., D < 0

9a2 32 b < 0

29a

b 0

32 > >


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Hence, a, b > 0

P (x) = 4x3+ 3ax2+ 2bx > 0 x > 0

P (x) is an increasing function on (0, 1)

P (0) < P (1)

Similarly we can prove P(x) is decreasing on (1, 0)

P(1) > P (0)

So we can conclude that

Max P (x) = P(1) and Min P(x) = P(0)

P(1) is not minimum but P (1) is the maximum of P.

68. Given, x y + 1 = 0 .(i)

and x = y2

dy

1 2y dx =

dy 1Slopeof givenline(i)

dx 2y = =

1

12y

=

1y

2 =

1y

2 =

21 1

x2 4

= =

Point (x, y)1 1

,4 2

=

The shortest distance is

1 11

34 2

1 1 4 2

+=

+

3 2

8=

69. The line isx 2 y 1 z 2

3 5 2

+= =

The direction ratios of the line are (3, 5, 2).

As the line lies in the plane x 3y z 0+ + =

We have (3) (1) + (5)(3) + 2() = 0

12 2 = 0


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412 AIEEE

= 6

Again (2, 1, 2) lies on the plane

2 + 3 + 2 + = 0

= 2 5 = 12 5 = 7

Hence, (, ) is (6, 7).

70. 4 novels, out of 6 novels and 1 dictionary out of 3 can b selected in 6 34 1C C waysThen 4 novels with one dictionary in the middle can be arranged in 4! ways.

Total ways of arrangement 6 34 1

C C 4! 1080= =

71. According to the condition

n3 9

14 10

n3 9 1

14 10 10

=

n4

103

n[log 4 log 3] log10

10 = 1

10 10

1n

log 4 log 3

72. The given lines are perpendicular to a common line, means they are in themselves parallel

The slope of rst line

( )2p p 1= +

The slope of second line( )22

2

2

p 1(p 1)

p 1

+= = +

+

On equating, we have

( ) ( )2 2p p 1 p 1+ = + p = 1

73. Let x A and x B

x A B

x A C (Since, A B = A C)

x C

B = C

Let x A and x B

x A B

x A C (Since, A B = A C)

x C

B = C


18/23


74. Given f(x) = x3+ 5x + 1

Now f(x) = 3x2+ 5 > 0, .x R

f(x) is strictly increasing function

f(x) is one-one function

Clearly f(x) is a continuous function and also increasing on R,

x xlim f(x) and limf(x)

= =

f(x) takes every value between and .

Thus, f(x) is onto function.

75. 2c x1y c e= differentiating w.r.t. x, we get

2c x

1 2y ' c c e= = c

2y .(i)

Again differentiating w.r.t. x y= c

2y .(ii)

From (i) and (ii) upon division

y ' y

y" y '=

yy = (y)2

Which is the desired differential equation of the family of curves.

76. ( )

1

na a 1 a 1 a 1 b 1 c 1b b 1 b 1 1 a 1 b 1 c 1

c c 1 c 1 a b c

+ + + + + +

+

( )na a 1 a 1 a 1 b 1 a

b b 1 b 1 1 b 1 b 1 b

c c 1 c 1 c 1 c 1 c

+ + +

= + +

+ +

( )n 1

2 3

a a 1 a 1 a 1 a a 1

b b 1 b 1 1 b 1 b b 1

c c 1 c 1 c 1 c c 1

C C

+

+ +

= + + +

+ +

( )( )n 2a a 1 a 1

1 1 b b 1 b 1

c c 1 c 1

+

+

= + +

+

This is equal to zero only if n + 2 is odd ie, n is an odd integer.

77. (8)2n (62)2n+1

= (64)n (62)2n+1


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414 AIEEE

= (63 + 1)n (63 1)2n+1

( ) ( ) ( )

( ) ( ) ( ) ( )

n n 1 n 2n n n n n

0 1 2 n 1 n

2n 1 n 2n 1 2n 12n 1 2n 1 2n 1 2n 1

0 1 2 2n 1

C 63 C 63 C 63 ..... C (63) C

C 63 C 63 C 63 .... 1 C

+ 2 ++ + + ++

= + + + + +

+ +

( ) ( ) ( )

( ) ( )

n 1 n 2 n 3n n n

0 1 2

2n 2n 12n 1 2n 1

0 1

63 C 63 C 63 C 63 .... 1

63 C 63 C 63 .... 1

+ +

= + + + +

+ +

63 some integral value + 2

82n (62)2n+1when divided by 9 leaves 2 as the remainder.

78. 2x xx 2x cot y 1 0 = (i)

Now, x = 1

1 2 coty 1 = 0

cot y = 0

y2

=

On differentiating Eq. (i), w.r.t. x, we get

( ) ( ) ( )2x x 2 xdy2x 1 logx 2 x cosec y cot y x 1 logx 0dx

+ + + =

At 1,2

( ) ( )1,

2

dy2 1 log1 2 1 1 0 0

dx

+ + =

1,2

dy2 2 0

dx

+ =

1,2

dy1

dx

=

79. The roots of bx2+ cx + a = 0 are imaginary means c2 4ab < 0

c2< 4ab

Again the coeff. of x2in

3b2x2+ 6bcx + 2c2is +ve, so the minimum value of the expression

( )( )( )

2 2 2 2 2 22

22

36b c 4 3b 2c 12b cc

12b4 3b

= = =

As c2< 4ab we have c2> 4ab

Thus theminimum value is 4ab.


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80. We have

2 3 4

2 6 10 14S 1 ..... ....(1)

3 3 3 3= + + + +

Multiplying both sides by

1

3 we get

2 3 4

1 1 2 6 10S ...... ....(2)

3 3 3 3 3= + + + +

Subtracting eqn. (2) from eqn. (1) we get

2 3 4

2 1 4 4 4S 1 .....

3 3 3 3 3= + + + + +

2 3 4

2 4 4 4 4S ......

3 3 3 3 3 = + + +

42 4 33S

13 3 21

3

= =

S = 3

81. Projection of a vector on coordinate axes are

x2 x

1, y

2 y

1, z

2 z

1

x2 x

1= 6, y

2 y

1= 3, z

2 z

1= 2

( ) ( ) ( )2 2 22 1 2 1 2 1x x y y z z + +

36 9 4 7= + + =

The DCs of the vector are6 3 2

, ,7 7 7

82. cos( ) + cos() + cos() =3

2

(cos cos + sin sin) + (cos cos + sin sin) + (cos cos + sin sin) =3

2

2(cos cos + cos cos + cos cos) + 2(sin sin + sin sin + sin sin) + 3 = 0

{cos2 + cos2 + cos2 + 2(cos cos + cos cos + cos cos)} + {sin2 + sin2 + sin2

+ 2(sin sin + sin sin + sin sin)} = 0 (cos + cos + cos)2+ (sin + sin + sin)2= 0

Which yields simultaneously

c os + cos + cos = 0 and sin + sin + sin = 0

83. Let A Sum of the digits is 8

B Product of the digits is 0

Then A = {08, 17, 26, 35, 44}

B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40}


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416 AIEEE

A B = {08}

( ) ( )P A B 1 / 50 1

P A / BP(B) 14 /50 14

= = =

84. Let (x, y) denotes the coordinates in A, B and C plane.

Then,( )( )

2 2

2 2

x 1 y 1

9x 1 y

+=

+ +

9x2+ 9y2 18x + 9 = x2+ y2+ 2x + 1

8x2+ 8y220x + 8 = 0

2 2 5x y x 1 02

+ + =

A, B, C lie on a circle with5

C ,0

4

85. The number are 1, 1+d, 1+2d, ., 1 + 100d.

The numbers are in A.P.

Then mean = 51stterm = 1 + 50d = x (says)

Mean deviation (M.D.)101

ii l

1x x

n ==

150d 49d 48d .... d 0 d 2d .... 50d

101= + + + + + + + + +

( )1 .2d 1 2 .... 50101

= + + +

1 50.51 50.51.2d. d

101 2 101= =

But M.D. = 255 (given)

50.51d 255

101 =

101 255 101 255d 10.1

50 51 2550

= = =

86. The given ellipse is2 2x y

14 1

+ =

So, A= (2, 0) and B = (0, 1)

If PQRS is the rectangle in which it is inscribed, then P = (2, 1).

Let2 2

2 2

x y1

a b+ = be the ellipse circumscribing the rectangle PQRS.


22/23


Then it passed through P (2, 1).

2 2

4 11 .....(a)

a b + =

Also, given that, it passes through (4, 0)

2

160 1

a + =

a2= 16

2 4b3

= [substituting a2= 16 in eqn(a)]

The required ellipse is2 2x y

1416

3

+ =

x2+ 12y2= 16

87. 4 4| z | zz z = +

4 4| z | z

z | z | +

4| z | 2

| z | +

|z|2 2|z| 4 0

( )( ) ( )( )| z | 5 1 | z | 1 5 0 +

1 5 | z | 5 1 +

88. The given circles are

S1x2+ y2+ 3x + 7y + 2p 5 = 0 .(i)

S2x2+ y2+ 2x + 2y p2= 0 .(ii)

equation of common chord PQ is S1 S

2= 0

L x + 5y + p2+ 2p 5 = 0 Equation of circle passing through P and Q is

S1+ L = 0

(x2+ y2+ 3x + 7y + 2p 5) + (x + 5y + p2+ 2p 5) = 0

As it passes through (1, 1), therefore,

(7 + 2p) + (2p + p2+ 1) = 0


23/23

418 AIEEE

( )22p 7

p 1

+ =

+

Which does not exist for p = 1.

89. We have la mb nc lmn a b c =

for scalars l, m, n.

Also a b c b c a c a b = =

(cyclic)

And a b c a c b =

(Interchange of any two vectors)

3u pv pw pu w qu 2w pv qu 0 =

2 2p u v w pq u v w 2q u v w 0 3 + =

( )2 23p pq 2q u v w 0 + =

As u vw

are non-coplanar, u v w 0

Hence, 3p2 pq + 2q2= 0, p, q R

As a quadratic in p, roots are real

q2 24q2 0 23q2 0

q2 0

q = 0And thus p = 0

Thus (p, q)(0, 0) is the only possibility.

90. Let0

I cot x dx ....(i)

=

( )0

I cot x dx

=

0cot x dx

= .(ii)

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

0 02I cot x dx cot x dx = +

( )0

1 dx

=

1 if x Zx x

0 if x Z

+ =

0x

= =

I2

=

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Past Year Papers AIEEE 2009

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