aieee paper 2005
TRANSCRIPT
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 1/23
1.
1.
2
2.
3.
3.
4.
4.
SOLUTION TO AIEEE-2005
M THEM TICS
If A' -A + I 0, then the inverse of A is
(1 )A+ I (2)A
{3)A- I (4)1-A
(4)ivenA2
- A+ I 'O
K 'A2- A-'A+ K - I K1 0 M u l ~ p l y m g A- 1 on both s1des)
=:.A-1 +A_, =OorA_ = 1-A.
If the cube roots of unity are 1, w, w then the roots of the equat onc;O(x-1)
3 +8=0,are
(1) -1,-1+2m,-1-2m' (2)-1,-1,-1
(3) -1 , 1- 2,,,, 1- 2w (4) -1 , 1 + 2w, 1 + 2w
(3) <0x -1 ) ' + 8 = 0 =:. (x -1 ) = {-2) (1) '
=>X-1=-2or-2mor-2m2 ror n ' -1 or 1 - wor 1 - 2w . -J
Let R '{(3, 3), (6, 6), (9, 9), (12, 12), 6 , ~ ~ 12), (3, 6)} be a relation onthe set A ' {3, 6, 9, 12} be a relation on I et , 6, 9, 12}. The relation is
(1) reflexive and transitive only re exive only(3) an equivalence relation reflexive and symmetric only(1)
Reflexive and transitive only.e.g. {3, 3), (6, 6), (9, 9), eflexive]
(3, 6), (6, 12), (3, 2.,. v [T rans i t i ve ] .
(1)2ab (2)ab
3 ) . / i b ~ • (4) ~(1)
ngle ABCD (2acas0)
= absin20(-orosH, I
'
' '
A(aoosB, bs' lf greatest rectangle is equal to
~ h e i o 2 0 • 1 . \\ J
' ' · """") : O(aoosB,
5. The differential equation representing the family of curves y' '2c( x + .JG , where c
> 0, is a parameter, is of order and degree as fallows:(1) arder1,degree2 (2)arder1,degree1
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 2/23
5.{3) order 1, degree 3
3)i =2c(x+..Jc) ... i)
2yy' 2c·1 or yy' c ... ii)
=> i ' 2yy' (x +.JY1On simplifying, we get
{4) order 2, degree 2
[on putting value of c from ii) in i)]
(y- 2xy•J 4yy ... iii)
Hence equation iii) is of order 1 and degree 3.
· [1 12 4 ]. hm --,sec -,+--,sec - , +-,sec 1 equals. - ~ n n n n n
1 11)
2sec1 2)
2cosec1
1{3) tan {4)
2tan1
6. (4)
7
{4) cosA: cosB: cosC
7.
= - - ~8. If in a frequently distribution, the mean and median are 21 and 22 respectively, then
tts mode is approximately
1) 22.0 2) 20.53) 25.5 4) 24.0
8. (4)Mode + 2Mean 3 Median
=>Mode"' 3 22-2 X 21= 66 - 42= 24
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 3/23
9
10.
10.
11.
11.
Let P be the pomt (1, 0) and Q a pam\ on the locus y "'8x. The locus of mid point of
PQis
(1)y ' -4x+2=0
(3)x '+4y+2=0
(1)P=(1,0)
Q (h, k) such that k' Bh
le t (q, )) be the midpoint of PQ
(2J i+4x+2=0
(4)-2--4y+2=0
u = h + l u , k + O
2 I ' 2u-1 =h 2fl=k.
{2 l)2 - 8 (2n- 1) => '- 4u -2
= > i - 4 x + 2 = 0 . 0If c is the mid P ( ) . ~ l of AB and Pis any point outsit _e AB, t h ~ r ; _ G(1) PA+PB=2PC (2) PA+PB=PC
(3) PA+PB+2PC=O (4) =0
(1)
- _ -··-PB+BC+CP=OAdding, we get
PA+PB+AC+BC+2CP =0
Since AC = -BC-··-
&CP=-PC _=:>PA + P B- 2PC = 0.
If the coefficients of rth, (r+y)m are in A.P., then
(1) m -m(4r-1) 1· , , ,_
(3) m - m(4r +(3)
Given me _
2mC ='
(4r+1)+4r"-2=0.
•
terms in the binomial expansion of (1 +equation
(2) m -m(4r+1) + 4r" + 2 = 0
(4) m -m(4r-1) + 4r" + 2 = 0
' " ' ' ' ' ' ' iPQR,, LR = . If tan ( ~ ) and tan ( ~ ) are the roots of
ax"+ bx + c = 0, a"" Othen(1 )a=b+c (2) c=a+b
(3)b=c {4 )b=a+c
12. (2)
tan(:) . t a n ( ~ ) a r e the roots of ax" +bx + c 0
' ( ~ l ' ( ~ H
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 4/23
13.
13.
14.
b
-a b a c=1 => - - = - - - = > -b=a-c ~ a a a
=a+b.
The system of equations
ux+y+z=u -1 ,
x + ay + z =a -1,
x+y+o.z=a-1
has no solution, i f« is L>_ +
~ J : : : ' , · u - 1 i ; J ~ ' ( ; ' V1 1 u
=u(u2
-1 ) -1 (u -1 )+ {
=> u-1)[u2+u -1 -1
=> u- 1 [u2
+a?lt[u
2 + 2u- u - 2]
{u-1)[u{u+2) ) =0
{a-1)= + 2 - a=-2, ; buleto<1
'''hi<oh tl e sum of the squares of the roots of the equation
= 0 assume tl e least value is
\ o - 2 • - a - 1 - 0
u+P=a-2=-(a+1)
+ p =(a+ 12
- 2 XP
=a -2a +6= (a-1) ' + 5
=>a= 1
(2) 0
(4) 2
15. If roots of the equation x2
- bx + c = 0 be two consectutive integers, then b2
- 4c
equals(1)-2 (2)3
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 5/23
15.
16.
16.
17.
17.
18.
3) 2(4)
Let a, a + 1 be roots
u u l = b
u a+l )=c
: b•-4c={2u+ t f -4u(u + 1 = 1.
4) 1
If the letters of word SACHIN are arranged in all possible ways and these words are
written out as in dictionary, then the word SACHIN appears at serial number1) 601 2) 6003) 603 4) 602
1)Alphabetical order is
0,C,H,I,N,S
No. of words starting with A - 51 Go. of words starting will1 C - 51
No. of words starting with H -51
No. of words starting with 1-51
No. of words starting with N - 51 0 +SACHIN -1
• 601.
GThe value of ' c .+ L '··c, is lj.(4)
c,•i' ·c,
=> C , - ~ [ C C + ~ C + C J
={ c: + C, ' + ~ ' C , ~ ~Y:,.c, + C,'
( C + C )0 + C + 'C4 ' ' ' '
=> c. =
, then which one of the following holds for all n ;, 1, by
' ~ 1 m • • ~ ' . ' " ' " ' ' ~ ' indunction- (n -1 1
+(n-1)1
2) A = 2 -'A- n - 1 I
(4) A = 2 -'A + n -1)1
the principle of mathematical induction 1) is true.
19. If the coefficient Of X in [ax' + b ~ ) r quals the coefficient Of X n[ax' - b ~ ) rthen a and b satisfy the relation
{1 )a-b=1 {2)a+b=1
'3) b ' 1 (4)ab= 1
19. 4)
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 6/23
20.
20.
21.
21.
22.
l- ' '] -- C,(••') ''(b',J',. 1 in the expansion ax +bx
c, (a) -' (b) ' (xt '_,,_,
=>-22-3r=7 =>r=5
: coefficient of x' ' C,(a) (br ...... 1)
Again T 1 in the expansion [ax- b:'r C, (ax) '' (- b: ' )'
c, a _, (-1)' {bt (x) '' (x) _Now11-3r=-7 =>3r=18 => r=6
coefficient of x ' Co a' x 1 {b)'
=> C, (a)' (b)' = C.a' x(b) '
=>ab=1. 0Let f: {-1, 1) 4 B, be a function defined by f(x) =tan '
2x, , t h ~ ~ t one-one
1 x ''ir.:1;, ,'(: ij ' ' 'h•lo1owol <'{o +
:::[-H] ~ ' /Given f(x) tan (
1::, for XE{-1, 1
clearty range off{x) = (-i· ~
(2) : I t
{4)- .::2
I + lzd > z1 and z, are collinear and are to the same side of origin;
~ · • - a r g z = 0 .
----;- and lwl = 1, then z lies on
z--i3
1) an ellipse
(3) a straight line
{3
(2) a circle
{4) a parabola.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 7/23
23.
23.
24.
s given w =----;- => 1w 1 I z 1 => dis1ance of z from origin and point
z - - i l z - - i l3 3
( 0, ± is same hence z lies on bisector of the line joining points {0, 0) and {0, 113).
Hence z lies on a straight line.
1+a x (1+b )x (1+c )x
1+b x (1+c )x thenf{x)isa ~fa2+ b + c
2=-2and f(x)= ( 1 + a ' ) ~
polynomial of degree{1 1
( 1 + a ) ~ (1+b )x 1 + c ' ~ ~{2)0 0{4)2 G3) 3
4)
1+(a +b +c +2)x (1+b )x {1+c )x
f x ) ~ 1+(a +b +c +2)x 1+b x {1+c )x , A p p l ~ , +C, + c,
1+(a +b +c +2)x (1+b )x 1+c x
J': : : : : ~ x i::::i: · · a + b + e 4 ~ C1 (1+b )x 1+c x . . . _ ,
0
f(x) 0
1 (1+b )x 1 ~ of (x)={x-1f
Hence degree .. .
The I t ~ x =a{cosO+ 0 s1nO), y =a{ smO- 0 cosO) at any pam '0' IS
_ + 0 with the x-axis2
at a constant distance from the oligln
Clealiy dy =tan 0 =>slope of normal=- cot 0,,Equation of normal at '0' is
y - a(sin 0-0 cos 0) cotO(x -a{cos 0 + 0 sin 0)
=>ysin 0- a sin" 0 +a 0 cos 0 sin 0 -x cos 0 +a cos" 0 +a 0 sin 0 cos 0
=o-xcosO+ysinO=a
Clearly this is an equation of straight line which is at a COI'lslant distance 'a' fromorigin.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 8/23
25. A funct1on is matched below aga1nst an mterval where IS supposed to beincreasing. Which of the following pairs is incorrect y matched?
Interval Function
{1) -oo,co) x ' -3x + 3x+3
(2) [2, oo) 2x3
- 3x' -12x + 6
{3)(.-..cc·±] 3x ' -2x+1
(4) (-m, -4] x + 6x' + 6
25. 3)Clearly function f{x) ' 3x ' - 2x + 1 is increasing when
f x ) = 6 x - 2 ~ 0 => XE[1/3,m)Hence (3) Is Incorrect.
26. Let a and ~ b e the distinct roots of ax' + bx + c ' 0, then
equal to
•1) 2 ( a - ~ )
•2 ( a - ~ )26. 1)
27.
28.
2
d i ~ r e n t i a b l e x 1 and l i m ~ f 1 +h)= 5, then f(1) equalsb ,, h
(2) 4
(4) 6
f(1+h)-f(1)lim ; AI; function is differentiable so tt. is continuous as it is given
. . hf(1+h)
lim -5 andhencef{1)=0h ,, h
f(1+h)Hence f(1) =lim -5
h ,, h
Hence (3) is the correct answer.
Let f be differentiable for all x. If f(1) 2 and f(x) 2 for x E [1, 6] , then
1 ) f 6 ) ~ 8 (2)f(6)<6
(3) f(6) < 5 (4) f(6) ' 5
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 9/23
28. (1)
29.
29.
30.
30.
Asf(1)=-2 &f(X)?:2 if X E (1,6)
Applying Lagrange s mean value theorem
f ( 6 ) ~ f ( 1 ) ~ t ( c ) ? : 25
=>f(6)?:10+f(1)
=>f(6)?:10-2
> f(6)?: 8.
3'--X .
8
• •, y L> · z= L;c where a, b, care in A.P. and [a[< 1, [b[<1, [c[< 1,
31. (4)
o 0 o·O
Aritl1metic e o m e t r i c Progression
1a
1b = -
y
(2) A.P.
{4) H.P.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 10/23
32.
32.
33.
33.
34.
. 1z=Lc =
' 1 ~ ca,b,carainA.P.
2b=a+c
2(1- )"' 1- +1-y ' y
2 1 1=y ' '=>x,y,zaremH.P
1c=1 - -
'
triangle ABC, then 2 (r + R) equals G1)b+c 2)a+b
3)a+b+c 4)c+a
(2)
y_ _ -COS X-COS 2 'Q
• [-;?2
. , / ( 1 ~ , < ' ) /· [ ~
=>4-i
~ ·cos•ox x o/- 4xy coso.
= 4 sin2u.
1 1 1'2p,a=2p,b=2p,b
p,, p2, p, are in H.P.
2t 2A 2t .- - -are m H.P.
' b '. HP
=>- , - , -a re1n .
' b '> a b, care in A.P.
> sinA, sinB, sinC are in A.P.
(2) A.P.
{4) H.P.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 11/23
1 j ' 2
35. If I, J 2' dx, 12
J2''dx, I ,= J2''dx and 14
J2'' dx then
35.
36.
36.
37.
37.
38.
38.
0 0 j '
(1)1,>1,
(3) I, ' 14(2)
'11 'J 2'' dx, 1 J2'' dx, 1 J2'' dx,
0
If 0 < x < 1, x2 > x
'=> J2''dx > J2''dx
0
(2) I,> 1 (4)1,>1,
The area enclosed between the curve y =log. x +e) and the coo < ; j t is(1)1 (2)2
(3)3 (4)4
(1) 0 r
Required area (OAB) J.ln(x +e):.lx~
• [x ln(x+e)- J- -•d•] 1.x+e 0
16
3
region bounded by the lines xare respectively the areas of these
: s, is
y - log x + 1 , then the solutton of the equation is
3 ) 1 o g ~ ) = c x(3),,,=y logy- log X+ 1
: : · H · , l ~ HPuty=vx
2 ) x l o g ~ ) = c y{ 4 ) 1 o g ~ ) = c y
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 12/23
39.
39.
dy xdv~ vdx dx
> v + x d v ~ v l o g v + 1 ),,,~ v l o g v
dv dx_
vlogv xput logv=z
1-dv =dz
dz dx~ - -
nz= lnx+ lnc
z =ex
logv=cx
l o g ( ~ ) ' +
The line parallel to the x-axis and passmg t rljri : sect1on of the lmes ax +2by + 3b 0 and bx- 2ay- 3a 0, Were (a, 'M(1) below the x-axis at a distance of ~ f r ~(2) below the x-axis at a d i s t a n c e ~ ~{3) above the x---axis at a ~ ~ m it
{4) above the x---axis ~ : ~ from 11(1)
ax+ 2by + 3b + ~ 3a) 0=> a+bA.)x+ 2 +3b-3A.a=O
a bi..=O A -
bx- 2ay-3a) 0
2a 3a3b-ax+ -Y - '
b b
l' '+ 3b+ b= O
Y(2b : 2a J - 3b : 3a )
-3(a +b ) -3Y - -
2(b +a ) 2
y = - ~ so it is 312 units below x-axis.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 13/23
40.
40.
41.
41.
A spherical iron ball 10 em radius s coated w1th a layer of ice of uniform thickness
than melts at a rate of 50 cm lmin. When the thickness of ice is 5 em, then the rate at
which the thickness of ice decreases, is
(1) -1-cmtmin
'3) -1-cmtmin
'2)
dv =50d1
4,.r d r- 50dt
dr 50- c- ;; ;c;o where r = 15
d t 4Jt(15)
1• .
Jr logx-f) l dxisequaltol(t+(logxf
(1) logx +C(logx) +1
3 ) ~ + C1+x
(4)
J (logx-1} x
(h(log•) )
· ) [ ~
(2) -1-cm/min
'4) ~ e m m i n
put logx = t =:- dx = e' dt
• '-+C ' +C1+t 1+(1ogx)
42. Let f R 4 R be a differentiable function having f (2) = 6, f {2) = ( ~ ) . Then
1 1 413
lim J equals•• • x -2
(1) 24
(3) 12(2) 36
(4) 16
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 14/23
42. (4)
43.
43.
44.
44.
1•1 'l i m ~ t••
0x-2
Applying L Hospital rule
~ ~ [ 4 f ( x ) ' f '(x)] =4f(2)3
f{2)
1=4 x6
3 =16.48
Let f {x) be a non-negative continuous function such that ~ e area b o u n d e ~curve y r (x), x-axls and the ordinates x - and x ~' • ( P • i o p ~ o o , p H / 2 j l l Thoo ( ~ ) ,,
4
r(11(: +J2-1) (2) ( ~ - J 2 + 1 -1
( 3 ) ( 1 - ~ - J 2 ) (4) ( 1 - ~ +(4) GGiven that 1(x)dx = ~ s i n ~ - i c o s ~ J 2 ~ ~Differentiating w. r. t . .V
' ( ~ H ~ ~ J ' ' ~ ~ ~
•
The loCI.ls of a po1nt~ ~
under the condition that the line y ax +IS
a
tangent to the hyr7 :,rb 1 is
(1) -/ {2) a circle(3) {4) a hyperbola
(4) •
x y'~ = 1 isa b
y o.x + is the tangent of hyperbola
' and a m b p'- b '
is a x - i= b which is hyperbola.
x+1 y-1 z - 2 r ;45. If the angle 0 between the line -
1- '
2 '-
2- and the plane 2 x - y + . {A. z + 4 =
0 is such that sin 0 i he value of J is
( 1 ) ~3
(2) -3
5
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 15/23
45.
46.
46.
47.
47.
48.
( 3 ) ~4
(1)Angle between line and normal to plane is
(4) --43
00, _:: -o =2
-2
+2./f. where 0 is angle between line & plane
2 3x.JS+' ..
2 fi 1=o; sinO= ~ ; ; - ' ' - c
J.JS+A 35o=;1..=-.
3
The angle between the lines 2x = 3y =- z and 6x = - y = - 4z is
1)0° (2)90
{3) 45° (4) 30°
2)
A ~ g l e between the lines 2x = 3y =- z & 6x = -y = -4z is ~ +S1nce a1a2+ b,b2 + c1c, = 0.
If the plane 2ax- 3ay + 4az + 6 = 0 passes t h r o u t : ~ , point of the line joining
the centres of the spheres l j ~ - -x' + + z' +6x- By -2z = 13 and
x' .; + z' -10x + 4y- 2z= 6, thenaes(1)-1 )3)-2 2
(3)Plane
2ax- 3ay + 4az + 6 = 0 p ~ the mid pomt of the centre of spheres
x + i + z 6 x - 6 y ~ 2 + ' i+z2-10x+4y-2z=8 respectively
centreofspheresare - ,1 , -2,1)
Mid pomt of centre s
Satisfymgthis ~ onofplane,weget
2 a - 3 a +4 a+6 'C J -2
The 1 ~ t w e e n the line f=2i-2]+3k+A. 1-]+4k) and the plane
"
; ' ~ t w e e o the line
( 2 ) ~3-/3
(4) ~3
f=2f-2]+3k+>.. 1-]+4k)and the planer 1+5]+k) =Sis
equation of plane is x + Sy + z = 5
: Distance of line from this plane= perpendicular distance of point (2, -2, 3) from the plane
2-10+3-5 10
i.e. ./1 +52 + 1 3./3 ·
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 16/23
49. For any vectoni , the value of (<i xi) +(<i jy + (<i k) is equal to
(1)38' (2) ii
(3)2S (4) 4ii'
49 (3)
50.
so.
Let ii=xi+y]+zk
ii.xi=z]-yk
=>(iixl) =y +zsimilarly (iix]) =x'+z'
and (iixk) =x +Y =>{iixf) =y +z
similarly ( ii ) 1) = x + z
and(iixk)'=x'+y'
=>(iixi)' +{iix]) +(iixk)' =2(x'+f'+z ) =2ii'. ~ +
If non-zero numbers a, b, care in H.P., then the 6 g ~ ~ y _ ~ = O' b '
passes through a fixed po1nt. That point IS
(1){-1,2)
(3)a,b,careinH.P.
2 lj> = 0
b
~ + r + = ob> < ~ ~ ~ - 1 r> r . y ~ - 21 2 1 '
always
51. trWngle is 1, 1) and the mid-points of two sides through this vertex
. ?:), then the centroid of the triangle is
( 2 ) ( ~ · ~ )(4) ( 1 : )
3 3
~ ~ ; ~ : : ~ ~ ; ~ ~ : : 1, 1) and midpoint of sidesis(-1,2)and{3,2)
Band C come out to be
(-3, 3) and {5, 3)
d. 1-3+5: centro1 1s3
,
=>(1,713)
1+3+3
3
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 17/23
52.
52.
53.
53.
54.
54.
If the circles + i + 2ax + cy + a = 0 and ¥ - + 3ax + d y- 1 "' 0 intersect in two
distinct points P and Q then the line Sx +by - a ' 0 passes through P and Q for
(1) exactly one value of a {2) no value of a
(3) infinitely many values of a {4) exactly two values of a
(2)S, = x' + i + 2ax + cy +a 0S2 =x'+Y' -3ax+dy-1 =0Equation of radical axis of 8 1 and s,
s,-s,,o=o; Sax+ (c-d)y+a + 1 0
Given that Sx + by- a " 0 passes through P and Q
a c - d a+1=0> o
1 b -ao=;a+1=-a2
a '+a+I=O
No real value of a
A circle touches the x-axis and also touches the circle w i ~ (0, 3) and radius2 The locus of the centre of the circle is
(1) an ellipse {2) a c ~ ·(3) a hyperbola (4) a p bol
(4) qquation of circle with centre (0, 3) and r a ~ · ·
x '+(y-3)2 4 .
Let locus of the variable circle is {a, p·:It touches x-a:ds.
: t equation {x-u) '+ {y- PlCircles touch externally
: J« '+(P 3)' =2+P
« '+ (p -3 } '=p '+
«' ' 10(p -112)
: Locus is ¥ - _t l land cuts the circle"¥ -+ y' = p' orthogonally,
= (2)2ax+2by-(a2- b
2+p
2)=0
= (4)2ax+2by-(a2
+b2
+p2) 0
p' orthogonally
X 0 + 2(-PJ X 0 'C
p
2
p'
Let equation of circle is"¥ -+ y '- 2«X- 2py + p2 = 0
It pass through (a, b) => a2
+ b '- 2«a- 2pb + p2
0
Locus : 2ax + 2by- (a + b2
+ p' ) 0
55. An ellipse has OB as semi minor axis, F and F' its focii and the angle FBF' is a right
angle. Then the eccentricity of the ellipse is
1 -
(2).. .J2 2
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 18/23
55.
56.
56.
57.
3 ) ~4
(1)
· LFBF' "'90"
: (.Ja'e' + b ) + (./a'e' + b ) = (2ae)"
=o 2(a2 e + b2
) ' 4a e
=o e2
' b2/a
2
Alsoe = 1-b2/a2 = 1 -e '
Vector al+a]+ck,
' '1 0 1 ' 0
' b
: .a ,b ,c are in G.P.
(4) _
f3
(2) no value of A
B(O, b)
{4) exactly two values of).
57.
0 , ,
0
value of Jc
00
only y {2) only xboth x andy {4) neither x nor y
58. (4)
.ii= 1 k, 6= xi+ j +(1-x)k and C= yl +X]+(1+x-y)k.
[•'+• '·')
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 19/23
59.
59.
60.
60.
61.
61.
j k
i i xC=x 1 ~ x =1(1+ x -x -x ) - ]{x+x -xy -y+xy )+k {x -y )
y x 1+x-y
a. bxC)= 1which does not depend on x and y.
lwo' ' such thatP A_u_BJ=i· P(ArtB)=± andP{A)=±·
stands for complement of event A. Then events A and Bare
~ ~ ~ ~ ; : ::and mutually exclusivei I but not independent
{3) independent but not equally likely{4) mutually exclusive and independen:
3)
P(AuB)=i· P A n B ) = ± and P A)=±
> P A u B)= 5 6 P A) = 3 4
Also P A u B)= P A) + P B)- P A r B)
> P B)= 516-314 + 1/4= 1/3
P A)P{B)=314-113=1/4 =P ArtB)
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 20/23
Hence A and 8 are independent but net equally likely.
62. A lizard, at an initial distance of 2 C ll behind an insect, moves from rest with anacceleration of 2 cmfs and pursues the insect which is crawling uniformly along a
straight line at a speed of 20 cmfs. Then the lizard will catch the insect after
(1)20s (2)1s(3)21s (4)24s
62. 3)
63.
63.
64.
64.
65.
2 t ' = 2 + 20t2
=>t=21.
eliminated and m we get
( f - f)n ' wm ' .2
A and B are two like ~ ? ' ~ ' ' ' ' ' " ' " ' ~B and is contained
through a distance
2H(1)
A-8
(3) 2(AH+B)
2)(A+B)=d=H
H
•
2)
moment H lies in the plane of A andA and B after combining is displaced
H
A•8
H(4)
A-8
R of two forces acting on a particle is at right angles to one of them and
is one third of the other force. The ratio of larger force to smaller one is
(2) 3 : .J2
' " 3Fcos0
F:3Fs in0
=> F' ' 2../2 F
F:F' : :3:2. . /2.
(4) 3:2../2
.
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 21/23
66.
67.
67.
66.
66.
69.
The sum of the series 1 + - _ , _ ,_ , _ , _ + ........ ad inf. is4.2 16.4 64.6
{1) e-1
rJ) e-1
. .
4)
putting x 1/2 we get
' cos' xThe value of J dx, a> 0 is
., 1+a
1) an:
3)
•
=o; (a+ b) ' + b2 + ab)
=o; 3a2 + 3b
2 + 2ab ' 0.
2) B+1
r4) B+1
. .
•
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 22/23
70. Let x1 , x2, ••• ,x, ben o b s e r v a ~ o n s such that l ;x, ' ~ 4 0 0 andl;x, ~ e o Then a
70.
possible value of n among the following is
{1 15
{3) 9
2)L · r ~ · ro = ; n ~ 1 6
{2) 16(4) 12
71. A particle is projected from a point 0 with velocity u at an angle of
h o r i l : o n t ; : ~ l When it is moving in a direction at right ; : ~ n g l e s to its ' ' ' ~velocity then is given by
( 1 ) ~ ( 2 ) ~3 2
{3) 2u (4)3
71. 4)
72.
73.
u cos 60 v cos 30
4v ' J3
•
{2) {6, ')
{4) [4, 5]
'•··· Min G.P., then lhe determinant
loga ''
loga.,, loga.,, is equal to
loga . , loga .•
c,,c,-c,two rows becomes identical
Answer: 0.
{2) 0{4) 2
74. A real valued function f{x) satisfies the functional equation f(x- y) ' f(x) f (y)- f (a- x)
f(a + y) where a is a given constant and f(O) ' 1 f {2a- x) is equal to
{1 -f(x) (2) f(x)
{3) f(a) + f (a- x) (4) f(-x)
8/12/2019 Aieee Paper 2005
http://slidepdf.com/reader/full/aieee-paper-2005 23/23
74. 1)
f(a- {x a))= f{a) f(x- a f O) f x)
=-f(x) [·: x=O, y=O, f(O)=f (O)-f (a) > f (a)=O > f(a)=OJ.