aieee paper 2005

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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


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

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9

10.

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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

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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

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16.

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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)

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20.

20.

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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.

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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.

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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

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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.

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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.

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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

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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.

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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

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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

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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 ·

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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

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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

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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.

[•'+• '·')

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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)

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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

.

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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

. .

•

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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)

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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.

aieee paper 2005

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