past year papers aieee 2009
TRANSCRIPT
-
8/10/2019 Past Year Papers AIEEE 2009
1/23
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
-
8/10/2019 Past Year Papers AIEEE 2009
2/23
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
-
8/10/2019 Past Year Papers AIEEE 2009
3/23
-
8/10/2019 Past Year Papers AIEEE 2009
4/23
399Past 5 Years Papers: Solved
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.
-
8/10/2019 Past Year Papers AIEEE 2009
5/23
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
=
-
8/10/2019 Past Year Papers AIEEE 2009
6/23
401Past 5 Years Papers: Solved
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
= = = =
-
8/10/2019 Past Year Papers AIEEE 2009
7/23
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 =
-
8/10/2019 Past Year Papers AIEEE 2009
8/23
403Past 5 Years Papers: Solved
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+ =
-
8/10/2019 Past Year Papers AIEEE 2009
9/23
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
+ = =
-
8/10/2019 Past Year Papers AIEEE 2009
10/23
405Past 5 Years Papers: Solved
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> > >
-
8/10/2019 Past Year Papers AIEEE 2009
11/23
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
= =
-
8/10/2019 Past Year Papers AIEEE 2009
12/23
407Past 5 Years Papers: Solved
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)
-
8/10/2019 Past Year Papers AIEEE 2009
13/23
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
-
8/10/2019 Past Year Papers AIEEE 2009
14/23
409Past 5 Years Papers: Solved
( )( )
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
-
8/10/2019 Past Year Papers AIEEE 2009
15/23
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 > >
-
8/10/2019 Past Year Papers AIEEE 2009
16/23
411Past 5 Years Papers: Solved
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
-
8/10/2019 Past Year Papers AIEEE 2009
17/23
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
-
8/10/2019 Past Year Papers AIEEE 2009
18/23
413Past 5 Years Papers: Solved
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
-
8/10/2019 Past Year Papers AIEEE 2009
19/23
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.
-
8/10/2019 Past Year Papers AIEEE 2009
20/23
415Past 5 Years Papers: Solved
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}
-
8/10/2019 Past Year Papers AIEEE 2009
21/23
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.
-
8/10/2019 Past Year Papers AIEEE 2009
22/23
417Past 5 Years Papers: Solved
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
-
8/10/2019 Past Year Papers AIEEE 2009
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
=