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AIEEECBSEENG03

1. A function f from the set of natural numbers to integers defined by

f (n) =

evenisnwhen,2

n

oddiswhen,2

1n

is

(A) oneone but not onto (B) onto but not oneone

(C) oneone and onto both () neither oneone nor onto

2. !et "1and "2be two roots of the e#uation "2$ a" $ b = %, " being com&le'.urther, assume that the origin, "1and "2form an e#uilateral triangle, then(A) a2= b (B) a2= 2b(C) a2= b () a2= *b

. +f " and are two non"ero com&le' numbers such that " = 1, and Arg (")

Arg () =2

, then " is e#ual to

(A) 1 (B) 1

(C) i () i

*. +f'

i1

i1

+

= 1, then

(A) ' = *n, where n is any &ositive integer

(B) ' = 2n, where n is any &ositive integer(C) ' = *n $ 1, where n is any &ositive integer() ' = 2n $ 1, where n is any &ositive integer

-. +f2

2

2

c1cc

b1bb

a1aa

+

+

+

= % and vectors (1, a, a2) (1, b, b2) and (1, c, c2) are non

co&lanar, then the &roduct abc e#uals(A) 2 (B) 1(C) 1 () %

. +f the system of linear e#uations' $ 2ay $ a" = %' $ by $ b" = %' $ *cy $ c" = %has a non"ero solution, then a, b, c(A) are in A. /. (B) are in 0./.(C) are in ./. () satisfy a $ 2b $ c = %

. +f the sum of the roots of the #uadratic e#uation a'2$ b' $ c = % is e#ual to the

sum of the s#uares of their reci&rocals, thena

b,

c

aand

b

care in

(A) arithmetic &rogression (B) geometric &rogression(C) harmonic &rogression () arithmeticgeometric&rogression

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3. 4he number of real solutions of the e#uation '2 ' $ 2 = % is(A) 2 (B) *(C) 1 ()

5. 4he value of 6a7 for which one root of the #uadratic e#uation(a2-a $ ) '2$ (a 1) ' $ 2 = % is twice as large as the other, is

(A)

2(B)

2

(C)

1()

1

+1%. +f A =

ab

baand A2=

, then

(A) = a2$ b2, = ab (B) = a2$ b2, = 2ab

(C) = a2$ b2, = a2b2 () = 2ab, = a2$ b2

11. A student is to answer 1% out of 1 #uestions in an e'amination such that hemust choose at least * from the first five #uestions. 4he number of choicesavailable to him is(A) 1*% (B) 15(C) 23% () *

12. 4he number of ways in which men and - women can dine at a round table if notwo women are to sit together is given by(A) 8 -8 (B) %

(C) -8 *8 () 8 -8

1. +f 1, , 2

are the cube roots of unity, then

=nn2

n2n

n2n

1

1

1

is e#ual to

(A) % (B) 1(C) () 2

1*. +f nCrdenotes the number of combinations of n things ta9en r at a time, then thee'&ression nCr$1$ nCr1$ 2 nCre#uals(A) n$2Cr (B) n$2Cr$1(C) n$1Cr () n$1Cr$1

1-. 4he number of integral terms in the e'&ansion of 2-3 )-( + is(A) 2 (B) (C) * () -

1. +f ' is &ositive, the first negative term in the e'&ansion of (1 $ ')2:-is(A) th term (B) -th term(C) 3th term () th term

1. 4he sum of the series*

1

2

1

21

1

+

;;; u&to is e#ual to

(A) 2 loge2 (B) log22 1

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(C) loge2 () loge

e

*

13. !et f (') be a &olynomial function of second degree. +f f (1) = f (1) and a, b, c

are in A. /., then f(a), f(b) and f(c) are in

(A) A./. (B) 0./.(C) . /. () arithmeticgeometric &rogression

15. +f '1, '2, 'and y1, y2, yare both in 0./. with the same common ratio, then the&oints ('1, y1) ('2, y2) and (', y)(A) lie on a straight line (B) lie on an elli&se(C) lie on a circle () are vertices of a triangle

2%. 4he sum of the radii of inscribed and circumscribed circles for an n sided regular&olygon of side a, is

(A) a cot

n

(B)

2

acot

n2

(C) a cot

n2

()*

acot

n2

21. +f in a triangle ABC a cos2

2

C$ c cos2

2

A=

2

b, then the sides a, b and c

(A) are in A./. (B) are in 0./.(C) are in ./. () satisfy a $ b = c

22. +n a triangle ABC, medians A and B< are drawn. +f A = *, AB =

and

AB< =

, then the area of the ABC is

(A)

3(B)

1

(C)

2()

*

2. 4he trigonometric e#uation sin1' = 2 sin1a, has a solution for

(A)2

1 a

2

1(B) all real values of a

(C) a 2

1() a

2

1

2*. 4he u&&er*

th &ortion of a vertical &ole subtends an angle tan1

-

at &oint in

the hori"ontal &lane through its foot and at a distance *% m from the foot. A&ossible height of the vertical &ole is(A) 2% m (B) *% m(C) % m () 3% m

2-. 4he real number ' when added to its inverse gives the minimum value of thesum at ' e#ual to(A) 2 (B) 1(C) 1 () 2

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2. +f f > ? ? satisfies f (' $ y) = f (') $ f (y), for all ', y ? and f (1) = , then

=

n

1r

)r(f is

(A)2

n(B)

2

)1n( +

(C) n (n $ 1) ()2

)1n(n +

2. +f f (') = 'n, then the value of f (1) 8n

)1(f)1(...

8

)1(f

82

)1(f

81

)1(f nn++

+

is

(A) 2n (B) 2n1(C) % () 1

23. omain of definition of the function f (') =2

'*

$ log1%(''), is

(A) (1, 2) (B) (1, %) (1, 2)

(C) (1, 2) (2, ) () (1, %) (1, 2) (2, )

25.

[ ]

[ ]2:'

'22

'tan1

'sin12

'tan1

lim

+

is

(A)

3

1(B) %

(C)2

1()

%. +f'

)'log()'log(lim

%'

+

= 9, the value of 9 is

(A) % (B)

1

(C)

2()

2

1. !et f (a) = g (a) = 9 and their nthderivatives fn(a), gn(a) e'ist and are not e#ual

for some n. urther if)'(f)'(g

)a(g)'(f)a(g)a(f)'(g)a(flim

a'

+

= *, then the value

of 9 is(A) * (B) 2(C) 1 () %

2. 4he function f (') = log (' $ 1'2 + ), is(A) an even function (B) an odd function(C) a &eriodic function () neither an even nor an odd function

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. +f f (') =

=

+

%',%

%','e '

1

'

1

then f (') is

(A) continuous as well as differentiable for all '(B) continuous for all ' but not differentiable at ' = %(C) neither differentiable nor continuous at ' = %() discontinuous everywhere

*. +f the function f (') = 2'5a'2$ 12a2' $ 1, where a @ %, attains its ma'imumand minimum at & and # res&ectively such that &2= #, then a e#uals(A) (B) 1

(C) 2 ()2

1

-. +f f (y) = ey, g (y) = y y @ % and (t) = t

%

f (t y) g (y) dy, then

(A) (t) = 1 et(1 $ t) (B) (t) = et(1 $ t)

(C) (t) = t et () (t) = t et

. +f f (a $ b ') = f ('), then b

a

' f (') d' is e#ual to

(A) +

b

a

d')'b(f2

ba(B)

+ b

a

d')'(f2

ba

(C)

b

a

d')'(f2

ab() +

+ b

a

d')'ba(f2

ba

. 4he value of

'sin'

dttsec

lim

2'

%

2

%'

is

(A) (B) 2(C) 1 () %

3. 4he value of the integral + = 1

%

' (1 ')nd' is

(A)1n

1

+(B)

2n

1

+

(C)1n

1

+

2n

1

+()

1n

1

+$

2n

1

+

5.-

n-

***

n n

n......21lim

n

n......21lim

++++

++++

is

(A)%1 (B) "ero

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(C)*

1()

-

1

*%. !etd'

d (') =

'

e 'sin, ' @ %. +f

'sin*

1

e'

d' = (9) (1), then one of the

&ossible values of 9, is(A) 1- (B) 1(C) () *

*1. 4he area of the region bounded by the curves y = ' 1 and y = ' is(A) 2 s# units (B) s# units(C) * s# units () s# units

*2. !et f (') be a function satisfying f(') = f (') with f (%) = 1 and g (') be a function

that satisfies f (') $ g (') = '2. 4hen the value of the integral 1

%

f (') g (') d', is

(A) e 2

-

2

e2 (B) e $

2

2

e2

(C) e 2

2

e2 () e $

2

-

2

e2+

*. 4he degree and order of the differential e#uation of the family of all &arabolaswhose a'is is 'a'is, are res&ectively(A) 2, 1 (B) 1, 2(C) , 2 () 2,

**. 4he solution of the differential e#uation (1 $ y2) $ (' ytan1

e

)d'dy = %, is

(A) (' 2) = 9 ytan1

e

(B) 2' ytan2

1

e

$ 9

(C) ' ytan1

e

= tan1y $ 9 () ' ytan2 1

e

= ytan1

e

$ 9

*-. +f the e#uation of the locus of a &oint e#uidistant from the &oints (a1, b1) and (a2,b2) is (a1a2) ' $ (b1b2) y $ c = %, then the value of 6c7 is

(A) )baba(2

1 21

21

22

22 + (B)

22

21

22

21 bbaa ++

(C) )bbaa(21 2

221

22

21 + () 22

22

21

21 baba +

*. !ocus of centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, bcos t) and (1, %), where t is a &arameter, is(A) (' 1)2$ (y)2= a2b2 (B) (' 1)2$ (y)2= a2$ b2

(C) (' $ 1)2$ (y)2= a2$ b2 () (' $ 1)2$ (y)2= a2b2

*. +f the &air of straight lines '22&'y y2= % and '22#'y y2= % be such thateach &air bisects the angle between the other &air, then(A) & = # (B) & = #

(C) = 1 () = 1

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*3. a s#uare of side a lies above the 'a'is and has one verte' at the origin. 4he

side &assing through the origin ma9es an angle (% *

) with the &ositive

direction of 'a'is. 4he e#uation of its diagonal not &assing through the origin is

(A) y (cos sin ) ' (sin cos ) = a

(B) y (cos $ sin ) $ ' (sin cos ) = a(C) y (cos $ sin ) $ ' (sin $ cos ) = a

() y (cos $ sin ) $ ' (cos sin ) = a

*5. +f the two circles (' 1)2$ (y )2= r2and '2$ y23' $ 2y $ 3 = % intersect intwo distinct &oints, then(A) 2 r 3 (B) r 2(C) r = 2 () r @ 2

-%. 4he lines 2' y = - and ' *y = are diameters of a circle having area as1-* s# units. 4hen the e#uation of the circle is

(A) '2

$ y2

$ 2' 2y = 2 (B) '2

$ y2

$ 2' 2y = *(C) '2$ y22' $ 2y = * () '2$ y22' $ 2y = 2

-1. 4he normal at the &oint (bt12, 2bt1) on a &arabola meets the &arabola again in the&oint (bt22, 2bt2), then

(A) t2= t11t

2(B) t2= t1$

1t

2

() t2= t11t

2() t2= t1$

1t

2

-2. 4he foci of the elli&se 2

22

by

1' + = 1 and the hy&erbola 2-131y1**'

22

= coincide.

4hen the value of b2is(A) 1 (B) -(C) () 5

-. A tetrahedron has vertices at (%, %, %), A (1, 2, 1), B (2, 1, ) and C (1, 1, 2).4hen the angle between the faces AB and ABC will be

(A) cos1

-

15(B) cos1

1

1

(C) %% () 5%%

-*. 4he radius of the circle in which the s&here '2$ y2$ "2$ 2' 2y *" 15 = % iscut by the &lane ' $ 2y $ 2" $ = % is(A) 1 (B) 2(C) () *

--. 4he lines9

*"

1

y

1

2'

=

=

and

1

-"

2

*y

9

1' =

=

are co&lanar if

(A) 9 = % or 1 (B) 9 = 1 or 1

(C) 9 = % or () 9 = or

-. 4he two lines ' = ay $ b, " = cy $ d and ' = ay $ b, " = cy $ dwill be&er&endicular, if and only if(A) aa$ bb$ cc$ 1 = % (B) aa$ bb$ cc= %

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(C) (a $ a) (b $ b) $ (c $ c) = % () aa$ cc$ 1 = %

-. 4he shortest distance from the &lane 12' $ *y $ " = 2 to the s&here '2$ y2

$ "2$ *' 2y " = 1-- is

(A) 2 (B) 11

1

*

(C) 1 () 5

-3. 4wo systems of rectangular a'es have the same origin. +f a &lane cuts them atdistances a, b, c and a, b, cfrom the origin, then

(A)

222222 c

1

b

1

a

1

c

1

b

1

a

1

+

+

+++ = % (B)

222222 c

1

b

1

a

1

c

1

b

1

a

1

+

++ = %

(C)

222222 c

1

b

1

a

1

c

1

b

1

a

1

+

= % ()

222222 c

1

b

1

a

1

c

1

b

1

a

1

++

= %

-5. c,b,a are vectors, such that %cba =++ , c,2b,1a == = , thenaccbba ++ is e#ual to

(A) % (B) (C) () 1

%. +f v,u and w are three nonco&lanar vectors, then ()wv()vu()wvu + e#uals

(A) % (B) wvu (C) vwu () wvu

1. Consider &oints A, B, C and with &osition vectors 91%D.i,9D*i ++

, 9C*DCiC + and 9C-DCiC- + res&ectively. 4hen ABC is a

(A) s#uare (B) rhombus(C) rectangle () &arallelogram but not a rhombus

2. 4he vectors 9C*iCAB += , and 9*D2i-AC += are the sides of a triangleABC. 4he length of the median through A is

(A) 13 (B) 2

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(C) () 233

. A &article acted on by constant forces 9CDCiC* + and 9CDCiC + is dis&laced

from the &oint 9CDC2iC ++ to the &oint 9CDC*iC- ++ . 4he total wor9 done by the

forces is

(A) 2% units (B) % units(C) *% units () -% units

*. !et DCiCv,DCiCu =+= and 9CDC2iCw ++= . +f nC is unit vector such that nCu = %

and nv = %, then nCw is e#ual to(A) % (B) 1(C) 2 ()

-. 4he median of a set of 5 distinct observations is 2%.-. +f each of the largest *observations of the set is increased by 2, then the median of the new set(A) is increased by 2 (B) is decreased by 2(C) is two times the original median () remains the same as that of theoriginal set

. +n an e'&eriment with 1- observations on ', then following results wereavailable>

2' = 23%, ' = 1%ne observation that was 2% was found to be wrong and was re&laced by thecorrect value %. 4hen the corrected variance is(A) 3.%% (B) 133.(C) 1. () 3.

. ive horses are in a race. Er. A selects two of the horses at random and bets on

them. 4he &robability that Er. A selected the winning horse is

(A)-

*(B)

-

(C)-

1()

-

2

3.

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%. 4he resultant of forces / and I is ? . +f I is doubled then ? is doubled. +f

the direction of I is reversed, then ? is again doubled. 4hen /2> I2> ?2is

(A) > 1 > 1 (B) 2 > > 2(C) 1 > 2 > () 2 > > 1

1. !et ?1 and ?2 res&ectively be the ma'imum ranges u& and down an inclined&lane and ? be the ma'imum range on the hori"ontal &lane. 4hen ? 1, ?, ?2 arein(A) arithmeticgeometric &rogression (B) A./.(C) 0./. () ./.

2. A cou&le is of moment 0 and the force forming the cou&le is / . +f / is turned

through a right angle, the moment of the cou&le thus formed is . +f instead,

the forces / are turned through an angle , then the moment of cou&le

becomes

(A) 0 sin cos (B) cos $ 0 sin

(C) 0 cos sin () sin 0 cos

. 4wo &articles start simultaneously from the same &oint and move along twostraight lines, one with uniform velocity uand the other from rest with uniform

acceleration f . !et be the angle between their directions of motion. 4he

relative velocity of the second &article with res&ect to the first is least after atime

(A)f

sinu (B)

u

cosf

(C) u sin ()

f

cosu

*. 4wo stones are &roDected from the to& of a cliff h meters high, with the sames&eed u so as to hit the ground at the same s&ot. +f one of the stones is &roDectedhori"ontally and the other is &roDected at an angle to the hori"ontal then tan e#uals

(A)gh

u2(B) 2g

h

u

(C) 2hg

u() u

gh

2

-. A body travels a distances s in t seconds. +t starts from rest and ends at rest. +nthe first &art of the Dourney, it moves with constant acceleration f and in thesecond &art with constant retardation r. 4he value of t is given by

(A) 2s

+r

1

f

1(B)

r

1

f

1

s2

+

(C) )rf(s2 + ()

+r

1

f

1s2

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Solutions

1. Clearly both one one and onto

Because if n is odd, values are set of all nonnegative integers and if n is aneven, values are set of all negative integers.ence, (C) is the correct answer.

2. "12$ "22"1"2= %("1$ "2)2"1"2= %a2= b.ence, (C) is the correct answer.

-.2

2

2

2

2

2

cc1

bb1

aa1

1cc

1bb

1aa

+ = %

(1 $ abc)

1cc

1bb

1aa

2

2

2

= %

abc = 1.ence, (B) is the correct answer

*.2

)i1(

i1

i1 2+=

+

= i

'

i1

i1

+

= i'

' = *n.ence, (A) is the correct answer.

. Coefficient determinant =cc*1

bb1

aa21

= %

b =ca

ac2

+.

ence, (C) is the correct answer

3. '2 ' $ 2 = %

(' 1) (' 2) = %

' = 1, 2.ence, (B) is the correct answer

. !et , be the roots

$ = 2211

+

$ =)(

222

++

2

2

c

ac2b

a

b =

2a2c = b (a2$ bc)

b

c,

a

b,

c

aare in ./.

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ence, (C) is the correct answer

1%. A =

ab

ba

A2=

ab

ba

ab

ba

=

++

22

22

baab2

ab2ba

= a2$ b2, = 2ab.ence, (B) is the correct answer.

5. = 2

=a-a

1a2 +

22

= .a-a

22 +

.a-a

1

)a-a(a

)1a(222

2

++=

+

a =

2.

ence, (A) is the correct answer

12. Clearly -8 8(A) is the correct answer

11. Jumber of choices = -C*3C$ -C-3C-= 1*% $ -.ence, (B) is the correct answer

1. =nn2n

n2n2n

n2nn2n

11

11

1

++

++

++

= %Kince, 1 $ n$ 2n= %, if n is not a multi&le of

4herefore, the roots are identical.

ence, (A) is the correct answer

1*. nCr$1$ nCr1$ nCr$ nCr= n$1Cr$1$ n$1Cr= n$2Cr$1.ence, (B) is the correct answer

1.*

1

2

1

21

1

+

;;;

= 1 *

1

1

1

2

1

2

1++ ;;;

= 1 2

+ .........

*1

1

21

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= 2

++ .........*

1

1

2

11 1

= 2 log 2 log e

= log

e

*.

ence, () is the correct answer.

1-. 0eneral term = 2-Cr( )2-rF(-)1:3Gr

rom integral terms, or should be 399 = % to 2.ence, (B) is the correct answer.

13. f (') = a'2$ b' $ cf (1) = a $ b $ cf (1) = a b $ c

a $ b $ c = a b $ c also 2b = a $ c

f(') = 2a' $ b = 2a'f(a) = 2a2

f(b) = 2ab

f(c) = 2ac

A/.ence, (A) is the correct answer.

15. ?esult (A) is correct answer.

2%. (B)

21. a2

b

2

Acos1c2

Ccos1 =

++

+

a $ c $ b = ba $ c = 2b.ence, (A) is the correct answer

2. f (1) = f (1 $ 1) = f (1) $ f (1)f (2) = 2

only f () =

=

n

1r)r(f = (1 $ 2 $ ;;; $ n)

= 2

)1n(n +.

2-. (B)

2. *

*2

'sin2

*

sin1(a)

*

2

1a

2

1.

ence, () is the correct answer

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2. !K = 1 8

)2n)(1n(n

82

)1n(n

81

n

+ $ ;;;

= 1 nC1$ nC2;;;= %.ence, (C) is the correct answer

%.

2

1'

1

'

1

lim%'

=

++

.

ence, (C) is the correct answer.

23. * '2%

' 2

'' @ %

' (' $ 1) (' 1) @ %.

ence () is the correct answer.

25.2

2:')'2(

2

'

**

)'sin1(2

'

*tan

lim

=2

1.


2. f (') = f (')ence, (B) is the correct answer.

1. sin ($ ) =*%

'

sin a =1*%

'

' = *%.ence, (B) is the correct answer

*%

':*

':*

tan1(:*)

*. f (') = % at ' = &, #&2$ 13a& $ 12a2= %#2$ 13a# $ 12a2= %f(') % at ' = &

and f(') @ % at ' = #.

%. A&&lying !. os&ital7s ?ule

)a(f)a(g

)a(f)a(g)a(g)a(flim

a2'

= *

))a(f)a(g())a(ff)a(g(9

= *

9 = *.

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ence, (A) is the correct answer.

. b

a

' f (') d'

= +b

a )'ba( f (a $ b ') d'.

ence, (B) is the correct answer.

. f (%)

f(% h) = 1

f(% $ h) = %

! ?.ence, (B) is the correct answer.

.

'sin'

)'tan(lim

2

%'

=

'

'sin'

)'tan(lim

2

2

%'

= 1.ence (C) is the correct answer.

3. 1

%

' (1 ')nd' = 1

%

n' (1 ')

= +

1

%

1nn )''( =2n

11n

1 ++ .


-. (t) = t

%

f (t y) f (y) dy

= t

%

f (y) f (t y) dy

= t

%

ye (t y) dy

= 't(1 $ t).ence, (B) is the correct answer.

*. Clearly f (') @ % for ' = 2a # = 2a % for ' = a & = a

or &2= # a = 2.ence, (C) is the correct answer.

*%. (') ='

'sin

e

= 'sine

' d' = (9) (1)

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= *

1

'sin

'

ed' = (9) (1)

= .*

1

(') d' = (9) (1)

(*) (1) = (9) (1)9 = *.ence, () is the correct answer.

*1. Clearly area = 2 2 2= s# units

(2,1)

(1,%)

(1,2)

*-. !et & (', y)(' a1)2$ (y b1)2= (' a2)2$ (y b2)2

(a1a2) ' $ (b1b2) y $2

1)aabb( 21

22

21

22 + = %.


*. ' =

1tsinbtcosa ++, y =

1tcosbtsina +

2

1'

$ y2=

5

ba 22+.

ence, (B) is the correct answer.

*.

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-2. 2

2

2

2

-

5

y

-

12

'

= 1

e1= *

-

ae2=1

b1

2

* =

b2= .ence, (C) is the correct answer.

-*. (C)

-

)

*

5. n& = *n = 2

# =2

1, & =

2

1

n = 3

& (' = 1) = 3C13

2

1

=21 .


*5. (' 1)2$ (y )2= r2

(' *)2$ (y $ 2)21 * $ 3 = %

(' *)2$ (y $ 2)2= 12.

. Kelect 2 out of -

=-

2.

ence, () is the correct answer.

-. % 2

'21

*

'1

1' +

+

+1

12' $ * $ ' $ 12' 1

% 1 ' 12

' 1

'

1

'

1.


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. Arg2

" =

" = 1

" = i or $ i.

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FIITJEE AIEEE 2004 (MATHEMATICS)

Important Instructions:

i) The test is of1

12

hours duration.

ii) The test consists of 75 questions.

iii) The maximum marks are 225.

iv) For each correct answer you will get 3 marks and for a wrong answer you will

get ! mark.

1. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3,4}. The relation R is(1) a !n"tion (2) rele#i$e(3) not s%&&etri" (4) transiti$e

2. The ran'e o the !n"tion # # 3(#) *

*= is(1) {1, 2, 3} (2) {1, 2, 3, 4, +}(3) {1, 2, 3, 4} (4) {1, 2, 3, 4, +, }

3. Let -, be "o&/le# n!&bers s!"h that - i0 = an ar' - = . Then ar' -

e!als

(1)4

/(2)

+

4

/

(3)3

4

/(4)

2

/

4. I - = # i % an13- / i= 0 , then

( )2 2

%#

/

/

5 607 8

9 :

0

is e!al to

(1) 1 (2) *2(3) 2 (4) *1

+. I 22- 1 - 1* = 0 , then - lies on

(1) the real a#is (2) an elli/se(3) a "ir"le (4) the i&a'inar% a#is.

. Let

1

A 1 .

1

*5 67 8

= *7 87 8*9 :

The onl% "orre"t state&ent abo!t the &atri# A is

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(1) A is a -ero &atri# (2) 2A I=(3) 1A* oes not e#ist (4) ( )A 1 I= * , here I is a !nit &atri#

. Let

1 1 1

A 2 1 3

1 1 1

*5 67

= *779 :

( )

4 2 2

1 ; +

1 2 3

5 67

= * a77 *9 :

. I ; is the in$erse o &atri# A, then is

(1) *2 (2) +(3) 2 (4) *1

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1+. The "oei"ient o the &ile ter& in the bino&ial e#/ansion in /oers o # o

( )4

1 #0a an o ( )

1 #* a is the sa&e i e!als

(1)+

3* (2)

3

+

(3) 31* (4) 1

3

1. The "oei"ient o n# in e#/ansion o ( ) ( )n

1 # 1 #0 * is

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

1 1 n* *

(3) ( ) ( )n 1 2

1 n 1*

* * (4) ( )n 1

1 n*

*

1. In

n nr r

1?

== an

n

n nr r

rt

== , then n

n

t

?is e!al to

(1)1n2

(2) 1n 12

*

(3) n 1 (4)2n 1

2

*

1

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21. Let , be s!"h that S * S 3. I sin0 sin=21

,+* an "os0 "os=

2

,+* ,

then the $al!e o "os2

a * bis

(1)3

13

* (2)3

13(3)

,

,+(4)

,

,+*

22. I 2 2 2 2 2 2 2 2! a "os b sin a sin b "os= 0 0 0 , then the ieren"e beteen the

&a#i&!& an &ini&!& $al!es o 2! is 'i$en b%

(1) ( )2 22 a b0 (2) 2 22 a b0(3) ( )

2a b0 (4) ( )

2a b*

23. The sies o a trian'le are sin, "osan 1 sin "os0 a a or so&e S S2/ .

Then the 'reatest an'le o the trian'le is

(1)o (2) >o

(3)12o (4) 1+o

24. A /erson stanin' on the ban o a ri$er obser$es that the an'le o ele$ation o

the to/ o a tree on the o//osite ban o the ri$er is oan hen he retires 4

&eter aa% ro& the tree the an'le o ele$ation be"o&es3o. The breath o theri$er is(1) 2 & (2) 3 &(3) 4 & (4) &

2+. I UR ?, eine b% (#) sin# 3"os# 1V = * 0 , is onto, then the inter$al o ? is(1) W, 3X (2) W*1, 1X(3) W, 1X (4) W*1, 3X

2. The 'ra/h o the !n"tion % = (#) is s%&&etri"al abo!t the line # = 2, then(1) (# 0 2)= (# 2) (2) (2 0 #) = (2 #)(3) (#) = (*#) (4) (#) = * (*#)

2. The o&ain o the !n"tion( )1

2

sin # 3(#)

> #

* *=

*is

(1) W2, 3X (2) W2, 3)(3) W1, 2X (4) W1, 2)

2

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2>. Let 1 tan#

(#) , # , # ,4# 4 2* / /L M= [ Z N O* / P Q

. I (#) is "ontin!o!s in , , then 2 4/ /L M 5 6

7N OP Q 9 :is

(1) 1 (2)1

2

(3)1

2* (4) *1

3. I % ...to% e# e

0 Y0= , # \ , then%

#is

(1)#

1 #0(2)

1

#

(3)1 #

#

*(4)

1 #

#

0

31. A /oint on the /arabola 2% 1 >

,< 2*5 6

79 :

(4)> >

,< 2

5 679 :

32. A !n"tion % = (#) has a se"on orer eri$ati$e (#) = (# 1). I its 'ra/h/asses thro!'h the /oint (2, 1) an at that /oint the tan'ent to the 'ra/h is % =3# +, then the !n"tion is

(1) ( )2

# 1* (2) ( )3

# 1*

(3) ( )3

# 10 (4) ( )2

# 10

33. The nor&al to the "!r$e # = a(1 0 "os), % = asinat GH ala%s /asses thro!'hthe i#e /oint(1) (a, ) (2) (, a)(3) (, ) (4) (a, a)

34. I 2a 0 3b 0 " =, then at least one root o the e!ation 2a# b# " 0 0 = lies inthe inter$al(1) (, 1) (2) (1, 2)(3) (2, 3) (4) (1, 3)

3+.r

n n

nr 1

1li& enVY =

is

(1) e (2) e 1(3) 1 e (4) e 0 1

3. I sin#

# A# ;lo'sin(# ) sin(# )

= 0 * a 0* a] , then $al!e o (A, ;) is

(1) (sin, "os) (2) ("os, sin)

(3) (* sin, "os) (4) (* "os, sin)

FIITJEE "td. #$%& 'ouse( &arvariya *ihar +,ear 'au- has /us Term.)( ,ew 0elhi !1( h 2141 5!42( 2115121( 2145 6!2( 215!56Fax 215!36

+

AIEEE*AER?*-5

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3. #"os# sin#*] is e!al to

(1) 1 #

lo' tan 2

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(3)1

lo'% #%

0 = (4) lo' % = #

4+. Let A (2, 3) an ;(2, 1) be $erti"es o a trian'le A;. I the "entroi o thistrian'le &o$es on the line 2# 0 3% = 1, then the lo"!s o the $erte# is the line(1) 2# 0 3% = > (2) 2# 3% = (3) 3# 0 2% = + (4) 3# 2% = 3

4. The e!ation o the strai'ht line /assin' thro!'h the /oint (4, 3) an &ain'inter"e/ts on the "o*orinate a#es hose s!& is 1 is

(1)%#

12 3

0 =* an %#

12 1

0 =**

(2)%#

12 3

* =* an%#

12 1

0 =**

(3)%#

12 3

0 = an %#

12 1

0 = (4)%#

12 3

* = an%#

12 1

0 =*

4. I the s!& o the slo/es o the lines 'i$en b% 2 2# 2"#% % * * = is o!r ti&es their

/ro!"t, then " has the $al!e(1) 1 (2) 1(3) 2 (4) 2

4

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(1) 2 2 (2b 3") 0 0 = (2) 2 2 (3b 2") 0 0 =

(3) 2 2 (2b 3") 0 * = (4) 2 2 (3b 2") 0 * =

+4. The e""entri"it% o an elli/se, ith its "entre at the ori'in, is1

2. I one o the

ire"tri"es is # = 4, then the e!ation o the elli/se is(1) 2 23# 4% 10 = (2) 2 23# 4% 120 =

(3) 2 24# 3% 120 = (4) 2 24# 3% 10 =

++. A line &aes the sa&e an'le , ith ea"h o the # an - a#is. I the an'le ,

hi"h it &aes ith %*a#is, is s!"h that 2 2sin 3sinb = , then 2"os e!als

(1)2

3(2)

1

+

(3)3

+(4)

2

+

+. Distan"e beteen to /arallel /lanes 2# 0 % 0 2- = < an 4# 0 2% 0 4- 0 + = is

(1)3

2(2)

+

2

(3)

2(4)

>

2

+. A line ith ire"tion "osines /ro/ortional to 2, 1, 2 &eets ea"h o the lines # = %0 a = - an# 0 a = 2% = 2-. The "o*orinates o ea"h o the /oint o interse"tion are 'i$en

b%(1) (3a, 3a, 3a), (a, a, a) (2) (3a, 2a, 3a), (a, a, a)(3) (3a, 2a, 3a), (a, a, 2a) (4) (2a, 3a, 3a), (2a, a, a)

+

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(1) al (2) bl(3) "l (4)

1. A /arti"le is a"te !/on b% "onstant or"es 4i 30 * an 3i 0 * hi"h is/la"e

it ro& a /oint i 2 30 0 to the /oint +i 4 0 0 . The or one in stanar !nits

b% the or"es is 'i$en b%(1) 4 (2) 3(3) 2+ (4) 1+

2. I a, b, " are non*"o/lanar $e"tors an is a real n!&ber, then the $e"tors

a 2b 3", b 4"0 0 l 0 an (2 1)"l * are non*"o/lanar or

(1) all $al!es o (2) all e#"e/t one $al!e o

(3) all e#"e/t to $al!es o (4) no $al!e o

3. Let !, $, be s!"h that! 1, $ 2, 3= = = . I the /roe"tion $ alon' ! is e!al

to that o alon' ! an $, are /er/eni"!lar to ea"h other then ! $ * 0 e!als

(1) 2 (2) (3) 14 (4) 14

4. Let a, b an " be non*-ero $e"tors s!"h that 1

(a b) " b " a3

c c = . I is the a"!te

an'le beteen the $e"tors b an" , then sin e!als

(1)1

3(2)

2

3

(3)2

3(4)

2 2

3

+. onsier the olloin' state&entsU(a) doe "an be "o&/!te ro& histo'ra&(b) deian is not ine/enent o "han'e o s"ale(") arian"e is ine/enent o "han'e o ori'in an s"ale.Khi"h o these is_are "orre"t(1) onl% (a) (2) onl% (b)(3) onl% (a) an (b) (4) (a), (b) an (")

. In a series o 2n obser$ations, hal o the& e!al a an re&ainin' hal e!al a.I the stanar e$iation o the obser$ations is 2, then ^a^ e!als

(1)1

n(2) 2

(3) 2 (4)2

n

. The /robabilit% that A s/eas tr!th is4

+, hile this /robabilit% or ; is

3

4. The

/robabilit% that the% "ontrai"t ea"h other hen ase to s/ea on a a"t is


>

AIEEE*AER?*-9

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(1)3

2(2)

1

+

(3)

2(4)

4

+

2+,

(3)12

2. Three or"es , g an R a"tin' alon' IA, I; an I, here I is the in"entre o a

A;, are in e!ilibri!&. Then Ug UR is

(1) A ; "os U"os U "os2 2 2 (2) A ; sin U sin U sin2 2 2

(3) A ;

se" U se" Use"2 2 2

(4)A ;

"ose" U"ose" U "ose"2 2 2

3. A /arti"le &o$es toars east ro& a /oint A to a /oint ; at the rate o 4 &_h

an then toars north ro& ; to at the rate o + &_h. I A; = 12 & an ;

= + &, then its a$era'e s/ee or its o!rne% ro& A to an res!ltant a$era'e

$elo"it% ire"t ro& A to are res/e"ti$el%

(1)1

4&_h an

13

4&_h (2)

13

4&_h an

1

4&_h


1

AIEEE*AER?*-10

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(3)1

>&_h an

13

>&_h (4)

13

>&_h an

1

>&_h

4. A $elo"it%1

4&_s is resol$e into to "o&/onents alon' A an ; &ain'

an'les 3 an 4+res/e"ti$el% ith the 'i$en $elo"it%. Then the "o&/onent

alon' ; is(1)

1

. I n is o then (n 1) is e$en s!& o o ter&s ( ) ( )2 22n 1 n n n 1n

2 2

* 0= 0 = .

2.2 4 e e

12 2 4

a * a0 a a a= 0 0 0 0..

2 4 e e1 .......

2 2 4

a * a0 a a a* = 0 0 0

/!t = 1, e 'et

( )2

e 1 1 1 1

2e 2 4

*= 0 0 ..

21. sin 0 sin =21

,+* an "os 0 "os =

2

,+* .

?!arin' an ain', e 'et

2 0 2 "os ( ) = 211

(+)

2 >"os

2 13a * b5 6=7 8

9 :

3"os

2 13

a * b *5 6=7 89 :

32 2 2/ a * b /5 6S S7

9 :g .

22. 2 2 2 2 2 2 2 2! a "os b sin a sin b "os= 0 0 0

=2 2 2 2 2 2 2 2a b a b a b b a

"os2 "os22 2 2 20 * 0 *

0 0 0

2 22 2 2 22 2 2 2a b a b! a b 2 "os 2

2 2

5 6 5 60 *= 0 0 * 7 8 7 8

9 : 9 :&in $al!e o 2 2 2! a b 2ab= 0 0

&a# $al!e o ( )2 2 2! 2 a b= 0

( )22 2

&a# &in! ! a b* = * .

23. reatest sie is 1 sin "os0 a a , b% a//l%in' "os r!le e 'et 'reatest an'le = 12.

24. tan3=h

4 b0 3h 4 b= 0

..(1)

tan= h_b h = 3 b .(2)

b = 2 &

h

b41

31 ,1


1+

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2+. 2 sin# 3"os# 2* j * j 1 sin# 3"os# 1 3* j * 0 jran'e o (#) is W1, 3X.Cen"e ? is W1, 3X.

2. I % = (#) is s%&&etri" abo!t the line # = 2 then (2 0 #) = (2 #).

2. 2> # * \ an 1 # 3 1* j * j # W2, 3)Z

2.#

4

1 tan# 1 tan# 1(#) li&

4# 4# 2/V

* *= B =*

* / * /

3. % .....

% e% e# e0 Y

00= # = % #e 0

ln# # = % % 1 1 #

1# # #

*= * = .

31. An% /oint be 2>t , >t2

5 679 :

m ierentiatin' %2= 1 1 1

2 ('i$en) t# % t 2

= = = B = .

oint is> >

,< 2

5 67

9 :32. (#) = (# 1) (#) = 3(# 1)20 "

an (2) = 3 " =

(#) = (# 1)30 an (2) = 1 =

(#) = (# 1)3.

33. Eli&inatin' , e 'et (# a)20 %2= a2.Cen"e nor&al ala%s /ass thro!'h (a, ).

34. Let (#) = 2a# b# "0 0 (#) =3 2a# b#

"# 3 2

0 0 0

( )3 21

(#) 2a# 3b# "#

= 0 0 0 , o (1) = () = , then a""orin' to RolleHs

theore&

(#) = 2a# b# " 0 0 = has at least one root in (, 1)

3+.rnn

nr 1

1li& e

nVY = =

1#

e # (e 1)= *]

3. !t # = t


1

AIEEE*AER?*-16

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sin( t)

t sin "ottt "os tsinta 0

= a 0 a] ] ]= ( )"os # sin ln sint "a * a 0 a 0

A = "os , ; sina = a

3. #"os# sin#*]

1 1 #2 "os #

4

=/5 607 8

9 :

] 1 se" # #42

/5 6= 07 89 :] =

1 # 3lo' tan

2

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2 2

(#%) 1%

%# % 0 =

1lo'%

#%* 0 = .

4+. I be (h, ) then "entroi is (h_3, ( 2)_3) it lies on 2# 0 3% = 1.lo"!s is 2# 0 3% = >.

4. %#

1a b

0 = here a 0 b = *1 an4 3

1a b

0 =

a = 2, b = *3 or a = *2, b = 1.

Cen"e % %# #

1an 12 3 2 1

* = 0 =*

.

4. &10 &2=2"

* an &1&2=

1

*

&10 &2= 4&1&2 ('i$en)" = 2.

4. Let the "ir"le be #20 %20 2'# 0 2% 0 " = " = 4 an it /asses thro!'h (a, b)

a20 b20 2'a 0 2b 0 4 = .Cen"e lo"!s o the "entre is 2a# 0 2b% (a20 b20 4) = .

+. Let the other en o ia&eter is (h, ) then e!ation o "ir"le is(# h)(# /) 0 (% )(% ) =

!t % = , sin"e #*a#is to!"hes the "ir"le#2 (h 0 /)# 0 (h/ 0 ) = (h 0 /)2= 4(h/ 0 ) (D = )

(# /)2= 4%.

+1. Interse"tion o 'i$en lines is the "entre o the "ir"le i.e. (1, 1)

ir"!&eren"e = 1rai!s r = +

e!ation o "ir"le is #20 %22# 0 2% 23 = .

+2. oints o interse"tion o line % = # ith #20 %22# = are (, ) an (1, 1)hen"e e!ation o "ir"le ha$in' en /oints o ia&eter (, ) an (1, 1) is#20 %2# % = .

+3. oints o interse"tion o 'i$en /arabolas are (, ) an (4a, 4a)e!ation o line /assin' thro!'h these /oints is % = #n "o&/arin' this line ith the 'i$en line 2b# 0 3"% 0 4 = , e 'et = an 2b 0 3" = (2b 0 3")20 2= .

+4. E!ation o ire"tri# is # = a_e = 4 a = 2

b2= a2(1 e2) b2= 3Cen"e e!ation o elli/se is 3#20 4%2= 12.


1

2

2 21 2 2

4!t t

'0 = .


21

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FIITJEELtd. ICES House, 29-A, Kalu Sarai , Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942

1

FIITJ

SOLUTION TO AIEEE-2005

MATHEMATICS

1. If A2 A + I = 0, then the inverse of A is(1) A + I (2) A(3) A I (4) I A

1. (4)Given A2 A + I = 0

A1A2 A1A + A1 I = A10 (Multiplying A1on both sides)A - I + A-1= 0 or A1= I A.

2. If the cube roots of unity are 1, , 2then the roots of the equation

(x 1)

3

+ 8 = 0, are(1) -1 , - 1 + 2, - 1 - 22 (2) -1 , -1, - 1(3) -1 , 1 - 2, 1 - 22 (4) -1 , 1 + 2, 1 + 22

2. (3)

(x 1)3+ 8 = 0 (x 1) = (-2) (1)1/3x 1 = -2 or -2or -22or n = -1 or 1 2or 1 22.

3. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation onthe set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is(1) reflexive and transitive only (2) reflexive only(3) an equivalence relation (4) reflexive and symmetric only

3. (1)Reflexive and transitive only.e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]

(3, 6), (6, 12), (3, 12) [Transitive].

4. Area of the greatest rectangle that can be inscribed in the ellipse2 2

2 2

x y1

a b+ = is

(1) 2ab (2) ab

(3) ab (4)a

b

4. (1)

Area of rectangle ABCD = (2acos)(2bsin) = 2absin2Area of greatest rectangle is equal to2ab

when sin2= 1.

(-acos, bsin)B

(-acos, -bsin)C D(acos, -bsin)

A(acos, bsin)

X

Y

5. The differential equation representing the family of curves y2= ( )2c x c+ , where c> 0, is a parameter, is of order and degree as follows:(1) order 1, degree 2 (2) order 1, degree 1

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(3) order 1, degree 3 (4) order 2, degree 25. (3)

y2= 2c(x + c) (i)2yy= 2c1 or yy= c (ii)y2= 2yy(x + yy ) [on putting value of c from (ii) in (i)]

On simplifying, we get(y 2xy)2= 4yy3 (iii)Hence equation (iii) is of order 1 and degree 3.

6. 2 2 22 2 2 2 2n

1 1 2 4 1lim sec sec .... sec 1

n n n n n

+ + + equals

(1)1

sec12

(2)1

cosec12

(3) tan1 (4)1

tan12

6. (4)

2 2 2 2

2 2 2 2 2 2n

1 1 2 4 3 9 1lim sec sec sec .... sec 1

nn n n n n n + + + +

is equal to

2 22 2

2 2 2n n

r r 1 r r lim sec lim sec

n nn n n =

Given limit is equal to value of integral1

2 2

0

x sec x dx

or1 1

2 2

0 0

1 12x sec x dx sec tdt

2 2= [put x2= t]

= ( )

1

0

1 1

tant tan12 2= .

7. ABC is a triangle. Forces P, Q, R

acting along IA, IB and IC respectively are in

equilibrium, where I is the incentre of ABC. Then P : Q : R is

(1) sinA : sin B : sinC (2)A B C

sin : sin : sin2 2 2

(3)A B C

cos : cos : cos2 2 2

(4) cosA : cosB : cosC

7. (3)Using Lamis Theorem

A B CP : Q :R cos : cos : cos2 2 2

= .

A

B C

I

P

Q R

8. If in a frequently distribution, the mean and median are 21 and 22 respectively, thenits mode is approximately(1) 22.0 (2) 20.5(3) 25.5 (4) 24.0

8. (4)Mode + 2Mean = 3 Median

Mode = 3 22 2 21= 66 42= 24.

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9. Let P be the point (1, 0) and Q a point on the locus y2= 8x. The locus of mid point ofPQ is(1) y2 4x + 2 = 0 (2) y2+ 4x + 2 = 0(3) x2+ 4y + 2 = 0 (4) x2 4y + 2 = 0

9. (1)P = (1, 0)Q = (h, k) such that k2= 8h

Let (, ) be the midpoint of PQh 1

2

+ = ,

k 0

2

+ =

2- 1 = h 2= k.(2)2= 8 (2- 1) 2= 4- 2y2 4x + 2 = 0.

10. If C is the mid point of AB and P is any point outside AB, then

(1) PA PB 2PC+ =

(2) PA PB PC+ =

(3) PA PB 2PC 0+ + =

(4) PA PB PC 0+ + =

10. (1)

PA AC CP 0+ + =

PB BC CP 0+ + =

Adding, we get

PA PB AC BC 2CP 0+ + + + =

Since AC BC=

& CP PC=

PA PB 2PC 0+ =

.

P

A CB

11. If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 +y)mare in A.P., then m and r satisfy the equation(1) m2 m(4r 1) + 4r2 2 = 0 (2) m2 m(4r+1) + 4r2+ 2 = 0(3) m2 m(4r + 1) + 4r2 2 = 0 (4) m2 m(4r 1) + 4r2+ 2 = 0

11. (3)

Given m m mr 1 r r 1C , C , C + are in A.P.m m m

r r 1 r 12 C C C += +

m m

r 1 r 1

m m

r r

C C2

C C += +

=

r m r

m r 1 r 1

+ + + m2 m (4r + 1) + 4r2 2 = 0.

12. In a triangle PQR, R =2

. If tan

P

2

and tanQ

2

are the roots of

ax2+ bx + c = 0, a 0 then(1) a = b + c (2) c = a + b(3) b = c (4) b = a + c

12. (2)

P Qtan , tan

2 2

are the roots of ax2+ bx + c = 0

P Q btan tan2 2 a

+ =

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P Q c

tan tan2 2 a

=

P Qtan tan

P Q2 2tan 1

P Q 2 21 tan tan2 2

+ = + =

b

a 1c

1a

=

b a c

a a a = b a c =

c = a + b.

13. The system of equations

x + y + z = - 1,x + y + z = - 1,

x + y + z = - 1has no solution, if is(1) -2 (2) either 2 or 1(3) not -2 (4) 1

13. (1)

x + y + z = - 1x + y + z = - 1x + y + z= - 1

1 1

1 1

1 1

=

= (2 1) 1(- 1) + 1(1 - )= (- 1) (+ 1) 1(- 1) 1(- 1)(- 1)[2+ - 1 1] = 0(- 1)[2+ - 2] = 0[2+ 2- - 2] = 0(- 1) [(+ 2) 1(+ 2)] = 0(- 1) = 0, + 2 = 0 = 2, 1; but 1.

14. The value of for which the sum of the squares of the roots of the equationx2 (a 2)x a 1 = 0 assume the least value is(1) 1 (2) 0

(3) 3 (4) 214. (1)

x2 (a 2)x a 1 = 0

+ = a 2= (a + 1)2+ 2= (+ )2- 2

= a2 2a + 6 = (a 1)2+ 5

a = 1.

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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(3) 2 (4) 115. (4)

Let , + 1 be roots+ + 1 = b(+ 1) = c

b2

4c = (2+ 1)2

- 4(+ 1) = 1.

16. If the letters of word SACHIN are arranged in all possible ways and these words arewritten out as in dictionary, then the word SACHIN appears at serial number(1) 601 (2) 600(3) 603 (4) 602

16. (1)Alphabetical order isA, C, H, I, N, SNo. of words starting with A 5!No. of words starting with C 5!No. of words starting with H 5!

No. of words starting with I 5!No. of words starting with N 5!

SACHIN 1601.

17. The value of 50C4+6

56 r

3r 1

C

= is

(1) 55C4 (2)55C3

(3) 56C3 (4)56C4

17. (4)

50C4+6

56 r

3r 1

C

=

50 55 54 53 52 51 50

4 3 3 3 3 3 3C C C C C C C + + + + + +

( )50 50 51 52 53 54 554 3 3 3 3 3 3C C C C C C C= + + + + + +

( )51 51 52 53 54 554 3 3 3 3 3C C C C C C+ + + + + 55C4+

55C3 =56C4.

18. If A =1 0

1 1

and I =1 0

0 1

, then which one of the following holds for all n 1, by

the principle of mathematical indunction

(1) An= nA (n 1)I (2) An= 2n-1A (n 1)I(3) An= nA + (n 1)I (4) An= 2n-1A + (n 1)I

18. (1)By the principle of mathematical induction (1) is true.

19. If the coefficient of x7in

11

2 1axbx

+

equals the coefficient of x-7in

11

2 1axbx

,

then a and b satisfy the relation(1) a b = 1 (2) a + b = 1

(3)a

b

= 1 (4) ab = 1

19. (4)

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Tr + 1in the expansion ( )11 r

11 r2 11 2

r

1 1ax C ax

bx bx

+ =

= 11Cr(a)11 r(b)-r(x)22 2r r

22 3r = 7 r = 5coefficient of x7= 11C5(a)

6(b)-5 (1)

Again Tr + 1in the expansion ( )11 r

11 r11

r2 2

1 1ax C ax

bx bx

=

= 11Cra11 r(-1)r(b)-r(x)-2r(x)11 - r

Now 11 3r = -7 3r = 18 r = 6coefficient of x-7= 11C6a

51 (b)-6

( ) ( ) ( )6 5 611 11 5

5 6C a b C a b

=

ab = 1.

20. Let f : (-1, 1) B, be a function defined by f(x) = 12

2xtan

1 x

, then f is both one-one

and onto when B is the interval

(1) 0,2

(2) 0,2

(3) ,2 2

(4) ,2 2

20. (4)

Given f(x) = 12

2xtan

1 x

for x(-1, 1)

clearly range of f(x) = ,

2 2

co-domain of function = B = ,2 2

.

21. If z1and z2are two non-zero complex numbers such that |z1+ z2| = |z1| + |z2| thenargz1 argz2is equal to

(1)2

(2) -

(3) 0 (4) -2

21. (3)

|z1+ z2| = |z1| + |z2| z1and z2are collinear and are to the same side of origin;hence arg z1 arg z2= 0.

22. If =z

1z i

3

and || = 1, then z lies on

(1) an ellipse (2) a circle(3) a straight line (4) a parabola.

22. (3)

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As givenz | z |

w | w |1 1

z i | z i |3 3

= =

= 1 distance of z from origin and point

10,

3

is same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).

Hence z lies on a straight line.

23. If a2+ b2+ c2= -2 and f(x) =

( ) ( )

( ) ( )

( ) ( )

2 2 2

2 2 2

2 2 2

1 a x 1 b x 1 c x

1 a x 1 b x 1 c x

1 a x 1 b x 1 c x

+ + +

+ + +

+ + +

then f(x) is a

polynomial of degree(1) 1 (2) 0(3) 3 (4) 2

23. (4)

( )( ) ( ) ( )( ) ( )

( ) ( )

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

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

1 a b c 2 x 1 b x 1 c x

+ + + + + +

= + + + + + +

+ + + + + +

, Applying C1C1+ C2+ C3

=

( ) ( )

( )

( )

2 2

2 2

2 2

1 1 b x 1 c x

1 1 b x 1 c x

1 1 b x 1 c x

+ +

+ +

+ +

2 2 2a b c 2 0+ + + =

f(x) =

( )2 2

0 x 1 0

0 1 x x 1

1 1 b x 1 c x

+ +

; Applying R1R1 R2, R2R2 R3

f(x) = (x 1)2Hence degree = 2.

24. The normal to the curve x = a(cos+ sin), y = a( sin- cos) at any point issuch that(1) it passes through the origin

(2) it makes angle2

+ with the x-axis

(3) it passes through a , a2

(4) it is at a constant distance from the origin24. (4)

Clearlydy

dx= tan slope of normal = - cot

Equation of normal at isy a(sin - cos ) = - cot (x a(cos + sin )y sin - a sin2+ a cos sin = -x cos + a cos2+ a sin cos x cos + y sin = aClearly this is an equation of straight line which is at a constant distance a from

origin.

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25. A function is matched below against an interval where it is supposed to beincreasing. Which of the following pairs is incorrectly matched?Interval Function

(1) (-, ) x3 3x2+ 3x + 3(2) [2, ) 2x3 3x2 12x + 6

(3) 1,3

3x2 2x + 1

(4) (- , -4] x3+ 6x2+ 625. (3)

Clearly function f(x) = 3x2 2x + 1 is increasing when

f(x) = 6x 2 0 x[1/3, )Hence (3) is incorrect.

26. Let and be the distinct roots of ax2+ bx + c = 0, then( )

( )

2

2x

1 cos ax bx clim

x

+ +

is

equal to(1) ( )

22a

2 (2) 0

(3) ( )2

2a

2 (4) ( )

21

2

26. (1)

Given limit =( ) ( )

( )

( ) ( )

( )

2

2 2x x

x x2sin a

21 cosa x xlim lim

x x

=

( )

( ) ( )

( ) ( )

( ) ( )2

2 22

2 2 22x

x x

sin a 2 a x x2lim

4x a x x

4

=

=( )

22a

2

.

27. Suppose f(x) is differentiable x = 1 and ( )h 0

1lim f 1 h 5

h+ = , then f(1) equals

(1) 3 (2) 4(3) 5 (4) 6

27. (3)

( ) ( ) ( )

h 0

f 1 h f 1f 1 lim

h

+ = ; As function is differentiable so it is continuous as it is given

that( )

h 0

f 1 hlim 5

h

+= and hence f(1) = 0

Hence f(1)( )

h 0

f 1 hlim 5

h

+= =

Hence (3) is the correct answer.

28. Let f be differentiable for all x. If f(1) = - 2 and f(x) 2 for x [1, 6] , then

(1) f(6) 8 (2) f(6) < 8(3) f(6) < 5 (4) f(6) = 5

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28. (1)

As f(1) = - 2 & f(x) 2 x [1, 6]Applying Lagranges mean value theorem

( ) ( )( )

f 6 f 1f c 2

5

=

f(6) 10 + f(1)f(6) 10 2f(6) 8.

29. If f is a real-valued differentiable function satisfying |f(x) f(y)| (x y)2, x, y R andf(0) = 0, then f(1) equals(1) -1 (2) 0(3) 2 (4) 1

29. (2)

f(x) =( ) ( )

h 0

f x h f xlim

h

+

( ) ( ) ( ) ( )2

h 0 h 0

f x h f x h| f x | lim lim

h h + =

|f(x)| 0 f(x) = 0 f(x) = constantAs f(0) = 0 f(1) = 0.

30. If x is so small that x3and higher powers of x may be neglected, then

( )

( )

33/ 2

1/ 2

11 x 1 x

2

1 x

+ +

may be approximated as

(1) 231 x8

(2) 233x x8

+

(3) 23

x8

(4) 2x 3

x2 8

30. (3)

(1 x)1/2 ( )2

23 3 3 1 11 x 1 x 1 3 x 3 2 x2 2 2 2 2

+ +

= (1 x)1/2 23

x8

= - 23

x8

.

31. If x = n n n

n 0 n 0 n 0

a , y b , z c

= = =

= = where a, b, c are in A.P. and |a| < 1, |b|

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z = n

n 0

1c

1 c

=

= c =

11

z

a, b, c are in A.P.2b = a + c

1 1 1

2 1 1 1y x y

= +

2 1 1

y x z= +

x, y, z are in H.P.

32. In a triangle ABC, let C =2

. If r is the inradius and R is the circumradius of the the

triangle ABC, then 2 (r + R) equals(1) b + c (2) a + b(3) a + b + c (4) c + a

32. (2)

2r + 2R = c +( )

( ) ( )( )

2a b c a b2ab

a ba b c a b c

+ + += = +

+ + + + ( since c2= a2+ b2).

33. If cos1x cos1y

2= , then 4x24xy cos + y2is equal to

(1) 2 sin 2 (2) 4(3) 4 sin2 (4) 4 sin2

33. (3)

cos-1x cos-1y

2 =

( )2

1 2xy ycos 1 x 12 4

+ =

2 2 2 21 xy 4 y 4x x ycos

2

+ + =

4 y2 4x2+ x2y2= 4 cos2+ x2y2 4xy cos4x2+ y2 4xy cos= 4 sin2.

34. If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in

H.P., then sin A, sin B, sin C are in(1) G.P. (2) A.P.

(3) Arithmetic Geometric Progression (4) H.P.34. (2)

= 1 2 31 1 1

p a p b p b2 2 2

= =

p1, p2, p3are in H.P.

2 2 2

, ,a b c

are in H.P.

1 1 1

, ,a b c

are in H.P

a, b, c are in A.P. sinA, sinB, sinC are in A.P.

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35. If I1=2 3

1 1

x x

2

0 0

2 dx, I 2 dx= , I3=2 3

2 2

x x

4

1 1

2 dx and I 2= dx then

(1) I2> I1 (2) I1> I2(3) I3= I4 (4) I3> I4

35. (2)

I1=2

1x

0

2 dx , I2=3

1x

0

2 dx , I3=2

1x

0

2 dx , I4=3

1x

0

2 dx

0 < x < 1, x2> x3

2

1

x

0

2 dx >3

1

x

0

2 dx

I1> I2.

36. The area enclosed between the curve y = loge(x + e) and the coordinate axes is(1) 1 (2) 2

(3) 3 (4) 436. (1)

Required area (OAB) = ( )0

1 e

ln x e dx

+

= ( )1

0

1x ln x e x dx

x e

+ +

= 1.

37. The parabolas y2= 4x and x2= 4y divide the square region bounded by the lines x =4, y = 4 and the coordinate axes. If S1, S2, S3are respectively the areas of theseparts numbered from top to bottom; then S1: S2: S3is

(1) 1 : 2 : 1 (2) 1 : 2 : 3(3) 2 : 1 : 2 (4) 1 : 1 : 137. (4)

y2= 4x and x2= 4y are symmetric about line y = x

area bounded between y2= 4x and y = x is ( )4

0

82 x x dx

3 =

2s

A =16

3 and

1 3s sA A= =

16

3

1s

A :2s

A :3s

A :: 1 : 1 : 1.

38. If x

dy

dx = y (log y log x + 1), then the solution of the equation is

(1) y logx

cxy

=

(2) x log

ycy

x

=

(3) logy

cxx

=

(4) log

xcy

y

=

38. (3)

xdyy

dx= (log y log x + 1)

dy y ylog 1

dx x x

= +

Put y = vx

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dy x dv

vdx dx

= +

( )xdv

v v log v 1dx

+ = +

xdv

vlogvdx = dv dx

v logv x =

put log v = z

1dv dz

v=

dz dx

z x =

ln z = ln x + ln cz = cx

log v = cxy

log cxx

=

.

39. The line parallel to the xaxis and passing through the intersection of the lines ax +2by + 3b = 0 and bx 2ay 3a = 0, where (a, b) (0, 0) is

(1) below the xaxis at a distance of3

2from it

(2) below the xaxis at a distance of2

3from it

(3) above the xaxis at a distance of3

2 from it

(4) above the xaxis at a distance of2

3from it

39. (1)

ax + 2by + 3b + (bx 2ay 3a) = 0(a + b)x + (2b 2a)y + 3b - 3a = 0a + b= 0 = -a/b

ax + 2by + 3b -a

b(bx 2ay 3a) = 0

ax + 2by + 3b ax +2 22a 3a

y 0b b

+ =

2 22a 3ay 2b 3b 0

b b

+ + + =

2 2 2 22b 2a 3b 3ay

b b

+ +=

( )( )

2 2

2 2

3 a b 3y

22 b a

+ = =

+

3y

2= so it is 3/2 units below x-axis.

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40. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thicknessthan melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate atwhich the thickness of ice decreases, is

(1)1

36cm/min (2)

1

18cm/min

(3) 154

cm/min (4) 56

cm/min

40. (2)

dv50

dt=

4r2dr

50dt

=

( )

2

dr 50

dt 4 15=

where r = 15

=1

16.

41.

2

2

(log x 1)dx

(1 (logx)

+ is equal to

(1)2

logxC

(logx) 1+

+ (2)

2

xC

x 1+

+

(3)x

2

xeC

1 x+

+ (4)

2

xC

(logx) 1+

+

41. (4)

( )( )( )

2

22

logx 1 dx1 logx

+

=( )( ) ( )( )

22 2

1 2logxdx

1 logx 1 logx

+ +

=

( )

t t

2 22

e 2t edt

1 t 1 t

+ +

put logx = t dx = etdt

( )t

2 22

1 2te dt1 t 1 t

+ +

=t

2

ec

1 t+

+=

( )2

xc

1 logx+

+

42. Let f : R R be a differentiable function having f (2) = 6, f (2) =1

48

. Then

f (x) 3

x 26

4tlim dt

x 2 equals

(1) 24 (2) 36(3) 12 (4) 18

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42. (4)( )f x 3

x 20

4tlim dt

x 2

Applying L Hospital rule

( ) ( )

2

x 2lim 4f x f x

= 4f(2)3

f(2)

= 4 631

48= 18.

43. Let f (x) be a nonnegative continuous function such that the area bounded by the

curve y = f (x), xaxis and the ordinates x =4

and x = >

4

is sin cos 24

+ +

. Then f

2

is

(1) 2 14

+

(2) 2 1

4

+

(3) 1 24

(4) 1 2

4

+

43. (4)

Given that ( )/ 4

f x dx sin cos 24

= + +

Differentiating w. r. t

f() = cos+ sin-4

sin+ 2

f 1 sin 2 1 22 4 2 4

= + = + .

44. The locus of a point P (, ) moving under the condition that the line y = x + is a

tangent to the hyperbola2 2

2 2

x y1

a b = is

(1) an ellipse (2) a circle(3) a parabola (4) a hyperbola

44. (4)

Tangent to the hyperbola2 2

2 2

x y1

a b = is

y = mx 2 2 2a m b Given that y = x + is the tangent of hyperbolam = and a2m2 b2= 2a22 b2= 2Locus is a2x2 y2= b2 which is hyperbola.

45. If the angle between the linex 1 y 1 z 2

1 2 2

+ = = and the plane 2x y + z + 4 =

0 is such that sin =1

3the value of is

(1)

5

3 (2)

3

5

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(3)3

4 (4)

4

3

45. (1)Angle between line and normal to plane is

2 2 2

cos 2 3 5

+

= + where

is angle between line & plane

sin=2 1

33 5

=

+

=5

3.

46. The angle between the lines 2x = 3y = z and 6x = y = 4z is(1) 00 (2) 900(3) 450 (4) 300

46. (2)Angle between the lines 2x = 3y = - z & 6x = -y = -4z is 90Since a1a2+ b1b2+ c1c2= 0.

47. If the plane 2ax 3ay + 4az + 6 = 0 passes through the midpoint of the line joiningthe centres of the spheres

x2+ y2+ z2+ 6x 8y 2z = 13 andx2+ y2+ z210x + 4y 2z = 8, then a equals(1) 1 (2) 1(3) 2 (4) 2

47. (3)Plane

2ax 3ay + 4az + 6 = 0 passes through the mid point of the centre of spheresx2+ y2+ z2+ 6x 8y 2z = 13 and x2+ y2+ z2 10x + 4y 2z = 8 respectivelycentre of spheres are (-3, 4, 1) & (5, - 2, 1)Mid point of centre is (1, 1, 1)Satisfying this in the equation of plane, we get

2a 3a + 4a + 6 = 0 a = -2.

48. The distance between the line r 2i 2 j 3k (i j 4k)= + + +

and the plane

r ( i 5 j k) 5 + + =

is

(1)10

9

(2)10

3 3

(3)3

10 (4)

10

3

48. (2)Distance between the line

( ) r 2i 2j 3k i j 4k= + + +

and the plane ( ) r i 5 j k + +

= 5 is

equation of plane is x + 5y + z = 5

Distance of line from this plane= perpendicular distance of point (2, -2, 3) from the plane

i.e.2

2 10 3 5 10

3 31 5 1

+ =

+ +

.

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49. For any vectora

, the value of 2 2 2 (a i) (a j) (a k) + +

is equal to

(1) 23a

(2) 2a

(3) 22a

(4) 24a

49. (3)

Let a xi yj zk= + +

a i zj yk =

( )2

2 2a i y z = +

similarly ( )2

2 2a j x z = +

and ( )2

2 2a k x y = +

( )2

2 2a i y z = +

similarly ( )2

2 2a j x z = +

and ( )2

2 2a k x y = +

( ) ( ) ( ) ( )2 2 2

2 2 2 a i a j a k 2 x y z + + = + +

= 2 2a

.

50. If non-zero numbers a, b, c are in H.P., then the straight linex y 1

0a b c

+ + = always

passes through a fixed point. That point is(1) (-1, 2) (2) (-1, -2)

(3) (1, -2) (4)1

1,2

50. (3)a, b, c are in H.P.

2 1 1 0b a c

=

x y 10

a b c+ + =

x y 1

1 2 1 = =

x = 1, y = -2

51. If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertexare (-1, 2) and (3, 2), then the centroid of the triangle is

(1)7

1,3

(2)1 7

,3 3

(3)7

1,3

(4)1 7

,3 3

51. (3)Vertex of triangle is (1, 1) and midpoint of sidesthrough this vertex is (-1, 2) and (3, 2)

vertex B and C come out to be(-3, 3) and (5, 3)

centroid is1 3 5 1 3 3

,3 3

+ + +

(1, 7/3)

A(1, 1)

(-1, 2)(3, 2)

CB

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52. If the circles x2+ y2+ 2ax + cy + a = 0 and x2+ y2 3ax + dy 1 = 0 intersect in two

distinct points P and Q then the line 5x + 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

52. (2)S1= x

2+ y2+ 2ax + cy + a = 0S2= x

2+ y2 3ax + dy 1 = 0Equation of radical axis of S1and S2S1 S2= 0

5ax + (c d)y + a + 1 = 0Given that 5x + by a = 0 passes through P and Q

a c d a 1

1 b a

+ = =

a + 1 = -a2a2+ a + 1 = 0No real value of a.

53. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius2. The locus of the centre of the circle is(1) an ellipse (2) a circle(3) a hyperbola (4) a parabola

53. (4)Equation of circle with centre (0, 3) and radius 2 isx2+ (y 3)2= 4.

Let locus of the variable circle is (, )It touches x-axis.

It equation (x - )2+ (y - )2= 2

Circles touch externally ( )

22 3 2 + = +

2+ (- 3)2= 2+ 4 + 42= 10(- 1/2)Locus is x2= 10(y 1/2) which is parabola.

(, )

54. If a circle passes through the point (a, b) and cuts the circle x2+ y2= p2orthogonally,then the equation of the locus of its centre is(1) x2+ y2 3ax 4by + (a2+ b2 p2) = 0 (2) 2ax + 2by (a2 b2+ p2) = 0(3) x2+ y2 2ax 3by + (a2 b2 p2) = 0 (4) 2ax + 2by (a2+ b2+ p2) = 0

54. (4)Let the centre be (, )It cut the circle x2+ y2= p2orthogonally

2(-) 0 + 2(-) 0 = c1 p2

c1= p2

Let equation of circle is x2+ y2- 2x - 2y + p2= 0It pass through (a, b) a2+ b2- 2a - 2b + p2= 0Locus 2ax + 2by (a2+ b2+ p2) = 0.

55. An ellipse has OB as semi minor axis, F and Fits focii and the angle FBFis a rightangle. Then the eccentricity of the ellipse is

(1)1

2 (2)1

2

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(3)1

4 (4)

1

3

55. (1)

FBF= 90o

( ) ( )

2 22 2 2 2 2 2

a e b a e b + + + = (2ae)2

2(a2e2+ b2) = 4a2e2e2= b2/a2Also e2= 1- b2/a2= 1 e2

B(0, b)

F(-ae, 0) O F(ae, 0)

21

2e 1, e2

= = .

56. Let a, b and c be distinct non-negative numbers. If the vectors ai aj ck, i k+ + + and ci cj bk+ + lie in a plane, then c is

(1) the Geometric Mean of a and b (2) the Arithmetic Mean of a and b

(3) equal to zero (4) the Harmonic Mean of a and b56. (1)

Vector ai aj ck+ + , i k+ and ci cj bk+ + are coplanar

a a c

1 0 1 0

c c b

= c2= ab

a, b, c are in G.P.

57. If a, b, c

are non-coplanar vectors and is a real number then

( )2

a b b c a b c b + = +

for(1) exactly one value of (2) no value of (3) exactly three values of (4) exactly two values of

57. (2)

( ) 2a b b c a b c b + = +

2

0 1 0 0

0 0 0 1 1

0 0 0 1 0

=

4= -1Hence no real value of .

58. Let ( ) a i k, b xi j 1 x k= = + +

and ( ) c yi xj 1 x y k= + + +

. Then a, b, c

depends on(1) only y (2) only x(3) both x and y (4) neither x nor y

58. (4) a i k=

, ( ) b xi j 1 x k= + +

and ( ) c yi xj 1 x y k= + + +

( )a b c a b c =

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i j k

b c x 1 1 x

y x 1 x y

=

+

= i (1 + x x x2) - j (x + x2- xy y + xy) + k (x2 y)

( )a. b c

= 1

which does not depend on x and y.

59. Three houses are available in a locality. Three persons apply for the houses. Eachapplies for one house without consulting others. The probability that all the threeapply for the same house is

(1)2

9 (2)

1

9

(3)8

9 (4)

7

9

59. (2)For a particular house being selected

Probability =1

3

Prob(all the persons apply for the same house) =1 1 1

33 3 3

=

1

9.

60. A random variable X has Poisson distribution with mean 2. Then P(X >1.5) equals

(1)2

2

e (2) 0

(3)2

31

e (4)

2

3

e

60. (3)

P(x = k) =k

ek!

P(x 2) = 1 P(x = 0) P(x = 1)

= 1 e- e- 1!

= 1 -2

3

e.

61. Let A and B be two events such that ( )1

P A B 6 = , ( )1

P A B 4 = and ( )1

P A 4= ,

where A stands for complement of event A. Then events A and B are(1) equally likely and mutually exclusive(2) equally likely but not independent(3) independent but not equally likely(4) mutually exclusive and independent

61. (3)

( ) 1P A B6

= , P(A B) =1

4and ( )

1P A

4=

P(A B) = 5/6 P(A) = 3/4Also P(A B) = P(A) + P(B) P(A B)P(B) = 5/6 3/4 + 1/4 = 1/3P(A) P(B) = 3/4 1/3 = 1/4 = P(A B)

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Hence A and B are independent but not equally likely.

62. A lizard, at an initial distance of 21 cm behind an insect, moves from rest with anacceleration of 2 cm/s2and pursues the insect which is crawling uniformly along astraight line at a speed of 20 cm/s. Then the lizard will catch the insect after(1) 20 s (2) 1 s(3) 21 s (4) 24 s

62. (3)

21 2t2

= 21 + 20t

t = 21.

63. Two points A and B move from rest along a straight line with constant acceleration f

and frespectively. If A takes m sec. more than B and describes n units more than Bin acquiring the same speed then

(1) (f - f)m2= ffn (2) (f + f)m2= ffn

(3) ( )21

f f m ff n2 + = (4) ( )21

f f n ff m2 = 63. (4)

v2= 2f(d + n) = 2fdv = f(t) = (m + t)feliminate d and m we get

(f- f)n = 21

ff m2

.

64. A and B are two like parallel forces. A couple of moment H lies in the plane of A andB and is contained with them. The resultant of A and B after combining is displacedthrough a distance

(1)2H

A B (2)

H

A B+

(3)( )

H

2 A B+ (4)

H

A B

64. (2)(A + B) = d = H

d =H

A B

+

.

65. The resultant R of two forces acting on a particle is at right angles to one of them and

its magnitude is one third of the other force. The ratio of larger force to smaller one is(1) 2 : 1 (2) 3 : 2

(3) 3 : 2 (4) 3 : 2 2

65. (4)

F= 3F cos F = 3F sin F= 2 2 F

F : F: : 3 : 2 2 .

3F

F

F

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66. The sum of the series1 1 1

1 .........4.2! 16.4! 64.6!

+ + + + ad inf. is

(1)e 1

e

(2)

e 1

e

+

(3)

e 1

2 e

(4)

e 1

2 e

+

66. (4)x x 2 4 6e e x x x

1 .......2 2! 4! 6!

+= + + + +

putting x = 1/2 we get

e 1

2 e

+.

67. The value of2

x

cos x

1 a

+ dx, a > 0, is

(1) a (2)2

(3)a

(4) 2

67. (2)2

2

x

0

cos xdx cos x dx

21 a

= =

+ .

68. The plane x + 2y z = 4 cuts the sphere x2+ y2+ z2 x + z 2 = 0 in a circle ofradius

(1) 3 (2) 1(3) 2 (4) 2

68. (2)

Perpendicular distance of centre1 1

, 0,2 2

from x + 2y 2 = 4

1 14

32 2

26

+ =

radius =5 3

12 2

= .

69. If the pair of lines ax2+ 2(a + b)xy + by2= 0 lie along diameters of a circle and dividethe circle into four sectors such that the area of one of the sectors is thrice the areaof another sector then(1) 3a2 10ab + 3b2= 0 (2) 3a2 2ab + 3b2= 0(3) 3a2+ 10ab + 3b2= 0 (4) 3a2+ 2ab + 3b2= 0

69. (4)

( )2

2 a b ab1

a b

+ =

+

(a + b)2= 4(a2+ b2+ ab)3a2+ 3b2+ 2ab = 0.

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70. Let x1, x2, ,xn be n observations such that2

ix 400= and ix 80= . Then a

possible value of n among the following is(1) 15 (2) 18(3) 9 (4) 12

70. (2)

22i ix x

n n

n 16.

71. A particle is projected from a point O with velocity u at an angle of 60o with thehorizontal. When it is moving in a direction at right angles to its direction at O, itsvelocity then is given by

(1)u

3 (2)

u

2

(3)2u

3 (4)

u

3

71. (4)u cos 60o= v cos 30o

v =4

3.

60o

30o

30o

72. If both the roots of the quadratic equation x2 2kx + k2+ k 5 = 0 are less than 5,

then k lies in the interval

(1) (5, 6] (2) (6, )(3) (-, 4) (4) [4, 5]

72. (3)

b5

2a

<

f(5) >0 k(-, 4).

73. If a1, a2, a3,, an, are in G.P., then the determinant

n n 1 n 2

n 3 n 4 n 5

n 6 n 7 n 8

loga loga loga

loga loga loga

loga loga loga

+ +

+ + +

+ + +

= is equal to

(1) 1 (2) 0(3) 4 (4) 2

73. (2)C1 C2, C2 C3two rows becomes identicalAnswer: 0.

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(0) = 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(0) f(x)

= -f(x) ( ) ( ) ( ) ( ) ( )2 2 2x 0, y 0, f 0 f 0 f a f a 0 f a 0 = = = = = .

75. If the equationn n 1

n n 1 1a x a x ...... a x 0+ + + = , a10, n 2, has a positive root x = , then the

equation ( )n 1 n 2n n 1 1na x n 1 a x ..... a 0

+ + + = has a positive root, which is

(1) greater than (2) smaller than (3) greater than or equal to (4) equal to

75. (2)

f(0) = 0, f() = 0f(k) = 0 for some k(0, ).

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FIITJ Solutions to AIEEE

2006

FIITJEELtd ., ICES House, 29 A, Kalu Sarai, Sarvapriya Vihar , New Delhi -110016, Ph 26515949, 26569493, Fax: 0 11-26 513942.

MATHEMATICS

PART A

1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC

with magnitudes 1AB

and 1AC

respectively is the force along AD

, where D is the

foot of the perpendicular from A onto BC. The magnitude of the resultant is

(1)2 2

2 2

AB AC

(AB) (AC)

+ (2)

(AB)(AC)

AB AC+

(3)1 1

AB AC+ (4)

1

AD

Ans. (4)

Sol: Magnitude of resultant

=

2 2 2 21 1 AB AC

AB AC AB AC

+ + =

BC BC 1

AB AC AD BC AD= = =

A B

C

D

2. Suppose a population A has 100 observations 101, 102, , 200, and anotherpopulation B has 100 observations 151, 152, , 250. If VA and VB represent the

variances of the two populations, respectively, then A

B

V

Vis

(1) 1 (2) 9/4

(3) 4/9 (4) 2/3

Ans. (1)

Sol:

2i2

x

d

n =

. (Here deviations are taken from the mean)

Since A and B both has 100 consecutive integers, therefore both have samestandard deviation and hence the variance.

AB

V1

V= ( )2iAs d is same in both the cases .

3. If the roots of the quadratic equation x2 + px + q = 0 are tan30 and tan15,

respectively then the value of 2 + q p is(3) 2 (2) 3(3) 0 (4) 1

Ans. (2)Sol: x2+ px + q = 0

tan 30+ tan 15= ptan 30tan 15= q

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tan 45=ptan 30 tan15

11 tan30 tan15 1 q

+ = =

p = 1 qq p = 1 2 + q p = 3.

4. The value of the integral,6

3

xdx

9 x x +is

(1) 1/2 (2) 3/2(3) 2 (4) 1

Ans. (2)

Sol: I =

6

3

xdx

9 x x +

I =

6

3

9 xdx

9 x x

+

2I =

6

3dx 3= I =

3

2 .

5. The number of values of x in the interval [0, 3] satisfying the equation2sin2x + 5sinx 3 = 0 is(1) 4 (2) 6(3) 1 (4) 2

Ans. (1)

Sol: 2 sin2x + 5 sin x 3 = 0(sin x + 3) (2 sin x 1) = 0

sin x =1

2 In (0, 3), x has 4 values

6. If ( ) ( )a b c a b c = , where a, b and c are any three vectors such that a b 0 ,

b c 0 , then a and c are(1) inclined at an angle of /3 between them(2) inclined at an angle of /6 between them(3) perpendicular(4) parallel

Ans. (4)

Sol:

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

( ) ( ) ( ) ( )a c b b c a a c b a b c =

( ) ( )a b c b c a = a c

7. Let W denote the words in the English dictionary. Define the relation R by :

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R = {(x, y) W W | the words x and y have at least one letter in common}. Then Ris(1) not reflexive, symmetric and transitive(2) reflexive, symmetric and not transitive(3) reflexive, symmetric and transitive(4) reflexive, not symmetric and transitive

Ans. (2)

Sol: Clearly (x, x) R x W. So, R is reflexive.Let (x, y) R, then (y, x) R as x and y have at least one letter in common. So, R issymmetric.But R is not transitive for exampleLet x = DELHI, y = DWARKA and z = PARK

then (x, y) R and (y, z) R but (x, z) R.

8. If A and B are square matrices of size n n such that A2B2= (A B) (A + B), thenwhich of the following will be always true ?(1) A = B(2) AB = BA(3) either of A or B is a zero matrix(4) either of A or B is an identity matrix

Ans. (2)

Sol: A2B2= (A B) (A + B)A2B2= A2+ AB BA B2AB = BA.

9. The value of10

k 1

2k 2ksin icos

11 11=

+

is

(1) i (2) 1(3) 1 (4) i

Ans. (4)

Sol:10 10 10

k 1 k 1 k 1

2k 2k 2k 2ksin icos sin i cos

11 11 11 11= = =

+ = +

= 0 + i (1) = i.

10. All the values of m for which both roots of the equations x22mx + m21 = 0 aregreater than 2 but less than 4, lie in the interval(1) 2 < m < 0 (2) m > 3

(3) 1 < m < 3 (4) 1 < m < 4

Ans. (3)

Sol: Equation x22mx + m21 = 0(x m)21 = 0(x m + 1) (x m 1) = 0x = m 1, m + 12 < m 1 and m + 1 < 4

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m > 1 and m < 31 < m < 3.

11. A particle has two velocities of equal magnitude inclined to each other at an angle .If one of them is halved, the angle between the other and the original resultant

velocity is bisected by the new resultant. Then is(1) 90 (2) 120(3) 45 (4) 60

Ans. (2)

Sol:

usin

2tanu4

u cos2

=

+

1 1

sin sin cos sin cos4 2 4 2 4

+ =

33

2sin sin 3sin 4sin4 4 4 4

= =

21

sin4 4

= 30

4

= or = 120.

u

u

/2

/4/4

R2 R1

u/2

12. At a telephone enquiry system the number of phone cells regarding relevant enquiryfollow Poisson distribution with an average of 5 phone calls during 10-minute timeintervals. The probability that there is at the most one phone call during a 10-minutetime period is

(1)e

6

5 (2)

5

6

(3)6

55 (4)

5

6

e

Ans. (4)

Sol: P (X = r) =m re m

r !

P (X 1) = P (X = 0) + P (X = 1)

= e5+ 5 e5=5

6

e.

13. A body falling from rest under gravity passes a certain point P. It was at a distance of400 m from P, 4s prior to passing through P. If g = 10 m/s2, then the height above the

point P from where the body began to fall is(1) 720 m (2) 900 m(3) 320 m (4) 680 m

Ans. (1)

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Sol: We have 21

h gt2

= and h + 400 = 21

g(t 4)2

+ .

Subtracting we get 400 = 8g + 4gt

t = 8 sec

1

h 10 64 320m

2

= =

Desired height = 320 + 400 = 720 m.

h

400m

Q(t)

P(t+4)

14.0

xf(sinx)dx

is equal to

(1)0

f(cosx)dx

(2)0

f(sinx)dx

(3)/ 2

0

f(sinx)dx2

(4)

/ 2

0

f(cosx)dx

Ans. (4)

Sol: I =

0 0

xf(sin x)dx ( x) f(sin x)dx

=

=

0

f(sin x) dx I

2I =

0

f(sinx)dx

I =

/ 2

0 0

f(sin x)dx f(sin x)dx

2

=

= / 2

0

f(cosx)dx

.

15. A straight line through the point A(3, 4) is such that its intercept between the axes isbisected at A. Its equation is

(1) x + y = 7 (2) 3x 4y + 7 = 0(3) 4x + 3y = 24 (4) 3x + 4y = 25

Ans. (3)Sol: The equation of axes is xy = 0

the equation of the line isx 4 y 312

2

+ = 4x + 3y = 24.

16. The two lines x = ay + b, z = cy + d; and x = ay + b, z = cy + dare perpendicular toeach other if

(1) aa+ cc= 1 (2) aa+ cc= 1

(3)a c

1a c

+ =

(4)a c

1a c

+ =

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Ans. (1)

Sol: Equation of linesx b z d

ya c

= =

x b z dy

a c

= =

Lines are perpendicular aa+ 1 + cc= 0.

17. The locus of the vertices of the family of parabolas3 2 2a x a x

y 2a3 2

= + is

(!)105

xy64

= (2)3

xy4

=

(3)35

xy16

= (4)64

xy105

=

Ans. (1)

Sol: Parabola: y =3 2 2a x a x

2a

3 2

+

Vertex: (, )

=2

3

a / 2 3

4a2a / 3

= , =

4 3

4

33

a a 1 84 2a a4 3 4 3

4a a433

+ + =

=35 a 35

3 a12 4 16

=

= 3 35 105

a4a 16 64

=

.

18. The values of a, for which the points A, B, C with position vectors 2i j k, i 3 j 5k + and ai 3 j k + respectively are the vertices of a right-angled

triangle with C2

= are

(1) 2 and 1 (2) 2 and 1(3) 2 and 1 (4) 2 and 1

Ans. (1)

Sol: BA i 2j 6k= +

CA (2 a)i 2 j= +

CB (1 a)i 6k=

CA CB

= 0 (2 a) (1 a) = 0a = 2, 1.

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19. ( ) ( )/ 2

3 2

3 / 2

x cos x 3 dx

+ + + is equal to

(1)4

32

(2)

4

32 2

+

(3) 2

(4) 14

Ans. (3)

Sol: I =

/ 2

3 2

3 / 2

(x ) cos (x 3 ) dx

+ + +

Put x + = t

I =

/ 2 / 2

3 2 2

/ 2 0

t cos t dt 2 cos t dt

+ =

=

/ 2

0

(1 cos 2t)dt 0

2

+ = +

.

20. If x is real, the maximum value of2

2

3x 9x 17

3x 9x 7

+ ++ +

is

(1) 1/4 (2) 41(3) 1 (4) 17/7

Ans. (2)

Sol:2

2

3x 9x 17y

3x 9x 7

+ +=

+ +

3x2(y 1) + 9x(y 1) + 7y 17 = 0

D 0 x is real( )( )281(y 1) 4x3 y 1 7y 17 0

(y 1) (y 41) 0 1 y 41.

21. In an ellipse, the distance between its foci is 6 and minor axis is 8. Then itseccentricity is

(1)3

5 (B)

1

2

(C)4

5 (D)

1

5

Ans. (1)

Sol: 2ae = 6 ae = 32b = 8 b = 4b2= a2(1 e2)16 = a2a2e2a2= 16 + 9 = 25a = 5

3 3e

a 5 = =

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2006


22. Let A =1 2

3 4

and B =a 0

0 b

, a , b N. Then

(1) there cannot exist any B such that AB = BA(2) there exist more than one but finite number of Bs such that AB = BA

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53664886 AIEEE Practice

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