(www.entrance-exam.net)-aieee queation papers for mathematics

8/12/2019 (Www.entrance-exam.net)-AIEEE Queation Papers for Mathematics

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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 function (2) reflexie(3) not s!""etric (4) transitie

2. The ran#e of the function$ x

x 3f(x) %−−= is

(1) {1, 2, 3} (2) {1, 2, 3, 4, &}(3) {1, 2, 3, 4} (4) {1, 2, 3, 4, &, '}

3. Let , be co"*lex nu"bers such that i+ = + an ar# = π. Then ar# e-uals

(1)4

π(2)

&

4

π

(3) 3

4π (4)

2π

4. f = x / i ! an1

3 * i-= + , then

( )2 2

!x

* -

* -

+ ÷

+

is e-ual to

(1) 1 (2) 02(3) 2 (4) 01

&. f 22 1 1− = + , then lies on

(1) the real axis (2) an elli*se(3) a circle (4) the i"a#inar! axis.

'. Let

+ + 1

A + 1 + .

1 + +

− ÷= − ÷ ÷−

The onl! correct state"ent about the "atrix A is

(1) A is a ero "atrix (2) 2 A =(3) 1 A− oes not exist (4) ( ) A 1 = − , here is a unit "atrix



$. Let

1 1 1

A 2 1 3

1 1 1

− ÷= − ÷ ÷

( )

4 2 2

1+ & +

1 2 3

÷= − α ÷ ÷−

. f is the inerse of "atrix A, then α is

(1) 02 (2) &(3) 2 (4) 01

. f 1 2 3 na , a , a , ....,a , .... are in .%., then the alue of the eter"inant

n n 1 n 2

n 3 n 4 n &

n ' n $ n

lo#a lo#a lo#a

lo#a lo#a lo#a

lo#a lo#a lo#a

+ +

+ + +

+ + +

, is

(1) + (2) 02(3) 2 (4) 1

. Let to nu"bers hae arith"etic "ean an #eo"etric "ean 4. Then these nu"bers arethe roots of the -uaratic e-uation

(1)2

x 1x 1' ++ + = (2)2

x 1x 1' +− − =(3) 2x 1x 1' ++ − = (4) 2x 1x 1' +− + =

1+. f (1 / *) is a root of -uaratic e-uation ( )2x *x 1 * ++ + − = , then its roots are

(1) +, 1 (2) 01, 2(3) +, 01 (4) 01, 1

11. Let ( ) 25(6) 1 3 & ... 26 1 3 6= + + + + − = + . Then hich of the folloin# is true7

(1) 5(1) is correct(2) %rinci*le of "athe"atical inuction can be use to *roe the for"ula

(3)5(6) 5(6 1)

⇒ +(4) 5(6) 5(6 1)⇒ +

12. 8o "an! a!s are there to arran#e the letters in the or AR9:; ith the oels inal*habetical orer7(1) 12+ (2) 4+(3) 3'+ (4) 24+

13. The nu"ber of a!s of istributin# ientical balls in 3 istinct boxes so that none of theboxes is e"*t! is

(1) & (2)

3<

(3)

3 (4) 21

14. f one root of the e-uation 2x *x 12 ++ + = is 4, hile the e-uation 2x *x - ++ + = has e-ual

roots, then the alue of -> is

(1)4

4(2) 4

(3) 3 (4) 12

FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949Fax : 2651394

2

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1&. The coefficient of the "ile ter" in the bino"ial ex*ansion in *oers of x of ( )4

1 x+ α an

of ( )'

1 x− α is the sa"e if α e-uals

(1)&

3− (2)

3

&

(3) 31+− (4) 1+

3

1'. The coefficient of nx in ex*ansion of ( ) ( )n

1 x 1 x+ − is

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

1 1 n− −

(3) ( ) ( )n 1 2

1 n 1−− − (4) ( )

n 11 n

−−

1$. f

n

n nr + r

15

<=

= ∑ an

n

n nr + r

r t

<=

= ∑ , thenn

n

t

5 is e-ual to

(1) 1n2

(2) 1 n 12

−

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

2

−

1. Let r T be the rth ter" of an A.%. hose first ter" is a an co""on ifference is . f for so"e

*ositie inte#ers ", n, " ≠ n, " n

1 1T an T

n "= = , then a / e-uals

(1) + (2) 1

(3)1

"n

(4)1 1

" n

+

1. The su" of the first n ter"s of the series 2 2 2 2 2 21 2 2 3 2 4 & 2 ' ...+ × + + × + + × + is( )

2n n 1

2

+

hen n is een. ?hen n is o the su" is

(1) ( )3n n 1

2

+(2)

( )2n n 1

2

+

(3) ( )

2n n 1

4

+(4)

( )2

n n 1

2

+

2+. The su" of series1 1 1

...2@ 4@ '@

+ + + is

(1)( )2e 1

2

−(2)

( )2

e 1

2e

−

(3)( )2e 1

2e

−(4)

( )2e 2

e

−


3

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21. Let α, β be such that π α 0 β 3π. f sinα B sinβ =21

'&− an cosα B cosβ =

2$

'&− , then the

alue of cos2

α − β is

(1)3

13+

− (2)3

13+

(3)'

'&(4)

'

'&−

22. f 2 2 2 2 2 2 2 2u a cos b sin a sin b cos= θ + θ + θ + θ , then the ifference beteen the

"axi"u" an "ini"u" alues of 2u is #ien b!

(1) ( )2 22 a b+ (2) 2 22 a b+

(3) ( )2

a b+ (4) ( )2

a b−

23. The sies of a trian#le are sinα, cosα an 1 sin cos+ α α for so"e + α 2π . Then the

#reatest an#le of the trian#le is

(1) '+o (2) +o

(3)12+o (4) 1&+o

24. A *erson stanin# on the banC of a rier obseres that the an#le of eleation of the to* of a

tree on the o**osite banC of the rier is '+o an hen he retires 4+ "eter aa! fro" the

tree the an#le of eleation beco"es3+o. The breath of the rier is

(1) 2+ " (2) 3+ "(3) 4+ " (4) '+ "

2&. f f D R 5, efine b! f(x) sin x 3 cos x 1→ = − + , is onto, then the interal of 5 is

(1) E+, 3F (2) E01, 1F(3) E+, 1F (4) E01, 3F

2'. The #ra*h of the function ! = f(x) is s!""etrical about the line x = 2, then(1) f(x B 2)= f(x / 2) (2) f(2 B x) = f(2 / x)(3) f(x) = f(0x) (4) f(x) = 0 f(0x)

2$. The o"ain of the function( )1

2

sin x 3f(x)

x

− −=

−

is

(1) E2, 3F (2) E2, 3)(3) E1, 2F (4) E1, 2)

2. f

2x

2

2x

a bli" 1 e

x x→∞

+ + = ÷

, then the alues of a an b, are

(1) a R, b R∈ ∈ (2) a = 1, b R∈

(3)a R, b 2∈ = (4) a = 1 an b = 2


4

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2. Let1 tan x

f(x) , x , x +,4x 4 2

− π π = ≠ ∈ − π . f f(x) is continuous in +, , then f

2 4

π π ÷

is

(1) 1 (2)1

2

(3) 1

2− (4) 01

3+. f ! ... to! ex e

+ ∞+= , x G +, then!

x is

(1)x

1 x+(2)

1

x

(3)1 x

x

−(4)

1 x

x

+

31. A *oint on the *arabola 2! 1x= at hich the orinate increases at tice the rate of the

abscissa is(1) (2, 4) (2) (2, 04)

(3)

, 2

− ÷

(4)

, 2

÷

32. A function ! = f(x) has a secon orer eriatie f ″(x) = '(x / 1). f its #ra*h *asses throu#h

the *oint (2, 1) an at that *oint the tan#ent to the #ra*h is ! = 3x / &, then the function is

(1) ( )2

x 1− (2) ( )3

x 1−

(3) ( )3

x 1+ (4) ( )2

x 1+

33. The nor"al to the cure x = a(1 B cosθ), ! = asinθ at θ> ala!s *asses throu#h the fixe*oint(1) (a, +) (2) (+, a)(3) (+, +) (4) (a, a)

34. f 2a B 3b B 'c =+, then at least one root of the e-uation 2ax bx c ++ + = lies in the interal

(1) (+, 1) (2) (1, 2)(3) (2, 3) (4) (1, 3)

3&.

r n

n

nr 1

1li" e

n→∞=

∑ is

(1) e (2) e / 1(3) 1 / e (4) e B 1

3'. f sinx

x Ax lo#sin(x ) <sin(x )

= + − α +− α∫ , then alue of (A, ) is

(1) (sinα, cosα) (2) (cosα, sinα)

(3) (0 sinα, cosα) (4) (0 cosα, sinα)

3$.x

cos x sin x−∫ is e-ual to


&

A:::0%A%:R50-5



(1)1 x

lo# tan <2 2

π − + ÷

(2)1 x

lo# cot <22

+ ÷

(3)1 x 3

lo# tan <2 2

π − + ÷

(4)1 x 3

lo# tan <2 2

π + + ÷

3. The alue of

3

2

2

H 1 x Hx

−

−∫ is

(1)2

3(2)

14

3

(3)$

3(4)

1

3

3. The alue of =

I 2 2

+

(sin x cos x)x

1 sin2x

π +

+∫ is

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

4+. f

I 2

+ +

xf(sin x)x A f(sin x)

π π

=∫ ∫ x, then A is

(1) + (2) π

(3)4

π(4) 2π

41. f f(x) =

x

x

e

1 e+ , 1 =

f(a)

f ( a)x#{x(1 x)}x

− −∫ an 2 =

f(a)

f ( a)#{x(1 x)}x

− −∫ then the alue of2

1

.

. is

(1) 2 (2) /3

(3) /1 (4) 1

42. The area of the re#ion boune b! the cures ! = Hx / 2H, x = 1, x = 3 an the x0axis is(1) 1 (2) 2

(3) 3 (4) 4

43. The ifferential e-uation for the fa"il! of cures 2 2x ! 2a! ++ − = , here a is an arbitrar!

constant is

(1) 2 22(x ! )! x!′− = (2) 2 22(x ! )! x!′+ =(3) 2 2(x ! )! 2x!′− = (4) 2 2(x ! )! 2x!′+ =

44. The solution of the ifferential e-uation ! x B (x B x2!) ! = + is

(1)1

<x!

− = (2)1

lo# ! <x!

− + =

(3)1

lo#! <x!

+ = (4) lo# ! = <x


'

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4&. Let A (2, /3) an (/2, 1) be ertices of a trian#le A<. f the centroi of this trian#le "oeson the line 2x B 3! = 1, then the locus of the ertex < is the line(1) 2x B 3! = (2) 2x / 3! = $(3) 3x B 2! = & (4) 3x / 2! = 3

4'. The e-uation of the strai#ht line *assin# throu#h the *oint (4, 3) an "aCin# interce*ts on

the co0orinate axes hose su" is /1 is

(1)!x

12 3

+ = − an!x

12 1

+ = −−

(2)!x

12 3

− = − an!x

12 1

+ = −−

(3)!x

12 3

+ = an!x

12 1

+ = (4)!x

12 3

− = an!x

12 1

+ =−

4$. f the su" of the slo*es of the lines #ien b! 2 2x 2cx! $! +− − = is four ti"es their *rouct,

then c has the alue(1) 1 (2) /1(3) 2 (4) /2

4. f one of the lines #ien b! 2 2'x x! 4c! +− + = is 3x B 4! = +, then c e-uals

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

4. f a circle *asses throu#h the *oint (a, b) an cuts the circle 2 2x ! 4+ = ortho#onall!, then

the locus of its centre is

(1) 2 22ax 2b! (a b 4) ++ + + + = (2) 2 22ax 2b! (a b 4) ++ − + + =

(3) 2 22ax 2b! (a b 4) +− + + + = (4) 2 22ax 2b! (a b 4) +− − + + =

&+. A ariable circle *asses throu#h the fixe *oint A (*, -) an touches x0axis. The locus of the

other en of the ia"eter throu#h A is

(1) 2(x *) 4-!− = (2) 2(x -) 4*!− =

(3) 2(! *) 4-x− = (4) 2(! -) 4*x− =

&1. f the lines 2x B 3! B 1 = + an 3x / ! / 4 = + lie alon# ia"eters of a circle of circu"ference

1+π, then the e-uation of the circle is

(1) 2 2x ! 2x 2! 23 ++ − + − = (2) 2 2x ! 2x 2! 23 ++ − − − =

(3) 2 2x ! 2x 2! 23 ++ + + − = (4) 2 2x ! 2x 2! 23 ++ + − − =

&2. The interce*t on the line ! = x b! the circle 2 2x ! 2x +

+ − = is A. :-uation of the circle on

A as a ia"eter is

(1) 2 2x ! x ! ++ − − = (2) 2 2x ! x ! ++ − + =

(3) 2 2x ! x ! ++ + + = (4) 2 2x ! x ! ++ + − =

&3. f a ≠ + an the line 2bx B 3c! B 4 = + *asses throu#h the *oints of intersection of the

*arabolas 2! 4ax= an 2x 4a!= , then

(1) 2 2 (2b 3c) ++ + = (2) 2 2 (3b 2c) ++ + =

(3) 2 2 (2b 3c) ++ − = (4) 2 2 (3b 2c) ++ − =


$

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&4. The eccentricit! of an elli*se, ith its centre at the ori#in, is1

2. f one of the irectrices is x =

4, then the e-uation of the elli*se is

(1) 2 23x 4! 1+ = (2) 2 23x 4! 12+ =

(3) 2 24x 3! 12+ = (4) 2 24x 3! 1+ =

&&. A line "aCes the sa"e an#le θ, ith each of the x an axis. f the an#le β, hich it "aCes

ith !0axis, is such that 2 2sin 3sinβ = θ , then 2cos θ e-uals

(1)2

3(2)

1

&

(3)3

&(4)

2

&

&'. 9istance beteen to *arallel *lanes 2x B ! B 2 = an 4x B 2! B 4 B & = + is

(1)3

2

(2)&

2

(3)$

2(4)

2

&$. A line ith irection cosines *ro*ortional to 2, 1, 2 "eets each of the lines x = ! B a = anx B a = 2! = 2. The co0orinates of each of the *oint of intersection are #ien 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)

&. f the strai#ht lines x = 1 B s, ! = /3 / λs, = 1 B λs an x =t

2, ! = 1 B t, = 2 / t ith

*ara"eters s an t res*ectiel!, are co0*lanar then λ e-uals(1) /2 (2) /1

(3) /1

2(4) +

&. The intersection of the s*heres 2 2 2x ! $x 2! 13+ + + − − = an

2 2 2x ! 3x 3! 4 + + − + + = is the sa"e as the intersection of one of the s*here an the

*lane(1) x / ! / = 1 (2) x / 2! / = 1(3) x / ! / 2 = 1 (4) 2x / ! / = 1

'+. Let a, b an c be three non0ero ectors such that no to of these are collinear. f the

ector a 2b+ is collinear ith c an b 3c+ is collinear ith a (λ bein# so"e non0ero

scalar) then a 2b 'c+ + e-uals

(1) aλ (2) bλ(3) cλ (4) +

'1. A *article is acte u*on b! constant forces J J J4i K 3C+ − an J J J3i K C+ − hich is*lace it fro" a

*oint J J Ji 2 K 3C+ + to the *oint J J J&i 4 K C+ + . The orC one in stanar units b! the forces is

#ien b!


A:::0%A%:R50-8



(1) 4+ (2) 3+(3) 2& (4) 1&

'2. f a, b, c are non0co*lanar ectors an λ is a real nu"ber, then the ectors

a 2b 3c, b 4c+ + λ + an (2 1)cλ − are non0co*lanar for

(1) all alues of λ (2) all exce*t one alue of λ(3) all exce*t to alues of λ (4) no alue of λ

'3. Let u, , be such that u 1, 2, 3= = = . f the *roKection alon# u is e-ual to that of

alon# u an , are *er*enicular to each other then u − + e-uals

(1) 2 (2) $

(3) 14 (4) 14

'4. Let a, b an c be non0ero ectors such that1

(a b) c b c a3

× × = . f θ is the acute an#le

beteen the ectors b an c , then sin θ e-uals

(1)1

3(2)

2

3

(3)2

3(4)

2 2

3

'&. <onsier the folloin# state"entsD

(a) oe can be co"*ute fro" histo#ra"

(b) eian is not ine*enent of chan#e of scale

(c) Mariance is ine*enent of chan#e of ori#in an scale.

?hich of these isIare correct7(1) onl! (a) (2) onl! (b)

(3) onl! (a) an (b) (4) (a), (b) an (c)

''. n a series of 2n obserations, half of the" e-ual a an re"ainin# half e-ual /a. f the

stanar eiation of the obserations is 2, then HaH e-uals

(1)1

n(2) 2

(3) 2 (4)2

n

'$. The *robabilit! that A s*eaCs truth is 4&

, hile this *robabilit! for is 34

. The *robabilit! that

the! contraict each other hen asCe to s*eaC on a fact is

(1)3

2+(2)

1

&

(3)$

2+(4)

4

&

'. A rano" ariable N has the *robabilit! istributionD

ND 1 2 3 4 & ' $

*(N)D +.1& +.23 +.12 +.1+ +.2+ +.+ +.+$ +.+&


A:::0%A%:R50-9



Oor the eents : = {N is a *ri"e nu"ber} an O = {N 4}, the *robabilit! % (: ∪ O) is

(1) +.$ (2) +.$$

(3) +.3& (4) +.&+

'. The "ean an the ariance of a bino"ial istribution are 4 an 2 res*ectiel!. Then the

*robabilit! of 2 successes is

(1)3$

2&'(2)

21

2&'

(3)12

2&'(4)

2

2&'

$+. ?ith to forces actin# at a *oint, the "axi"u" effect is obtaine hen their resultant is 4;.

f the! act at ri#ht an#les, then their resultant is 3;. Then the forces are

(1)(2 2); an (2 2);+ − (2) (2 3); an (2 3);+ −

(3) 1 1

2 2 ; an 2 2 ;2 2

+ − ÷ ÷

(4)1 1

2 3 ; an 2 3 ;2 2

+ − ÷ ÷

$1. n a ri#ht an#le ∆ A<, ∠ A = +° an sies a, b, c are res*ectiel!, & c", 4 c" an 3 c". f a

force O has "o"ents +, an 1' in ; c". units res*ectiel! about ertices A, an <,

then "a#nitue of O is

(1) 3 (2) 4

(3) & (4)

$2. Three forces %, P an R actin# alon# A, an <, here is the incentre of a ∆ A<, are

in e-uilibriu". Then % D P D R is

(1) A <

cos D cos D cos2 2 2 (2) A <

sin D sin D sin2 2 2

(3) A <

sec D sec D sec2 2 2

(4) A <

cosec D cosec D cosec2 2 2

$3. A *article "oes toars east fro" a *oint A to a *oint at the rate of 4 C"Ih an then

toars north fro" to < at the rate of & C"Ih. f A = 12 C" an < = & C", then its

aera#e s*ee for its Kourne! fro" A to < an resultant aera#e elocit! irect fro" A to <

are res*ectiel!

(1)1$

4C"Ih an

13

4 C"Ih (2)

13

4C"Ih an

1$

4 C"Ih

(3) 1$

C"Ih an 13

C"Ih (4) 13

C"Ih an 1$

C"Ih

$4. A elocit!1

4 "Is is resole into to co"*onents alon# QA an Q "aCin# an#les 3+° an

4&° res*ectiel! ith the #ien elocit!. Then the co"*onent alon# Q is

(1)1

"Is (2)

1( 3 1)

4− "Is

(3)1

4"Is (4)

1( ' 2)

− "Is


1+

A:::0%A%:R50-10



$&. f t1 an t2 are the ti"es of fli#ht of to *articles hain# the sa"e initial elocit! u an ran#e

R on the horiontal, then 2 21 2t t+ is e-ual to

(1)2u

#(2)

2

2

4u

#

(3)

2u

2# (4) 1


11

A:::0%A%:R50-11




ANSWERS

1. 3 1'. 2 31. 4 4'. 4 '1. 12. 1 1$. 1 32. 2 4$. 3 '2. 33. 3 1. 1 33. 1 4. 4 '3. 34. 2 1. 2 34. 1 4. 2 '4. 4&. 4 2+. 2 3&. 2 &+. 1 '&. 3'. 2 21. 1 3'. 2 &1. 1 ''. 3$. 2 22. 4 3$. 4 &2. 1 '$. 3. 1 23. 3 3. 1 &3. 1 '. 2. 4 24. 1 3. 3 &4. 2 '. 41+. 3 2&. 4 4+. 2 &&. 3 $+. 311. 4 2'. 2 41. 1 &'. 3 $1. 312. 3 2$. 2 42. 1 &$. 2 $2. 113. 4 2. 2 43. 3 &. 1 $3. 114. 1 2. 3 44. 2 &. 4 $4. 41&. 3 3+. 3 4&. 1 '+. 4 $&. 2


12

A:::0%A%:R50-12




SOLUTIONS

1. (2, 3) ∈ R but (3, 2) ∉ R.

8ence R is not s!""etric.

2.$ x

x 3f(x) %−−=

$ x + x $− ≥ ⇒ ≤x 3 + x 3− ≥ ⇒ ≥ ,

an $ x x 3 x &− ≥ − ⇒ ≤⇒ 3 x &≤ ≤ ⇒ x = 3, 4, & ⇒ Ran#e is {1, 2, 3}.

3. 8ere ω =

i ⇒ ar#

.

i

= π ÷

⇒ 2 ar#() / ar#(i) = π ⇒ ar#() =3

4

π.

4. ( ) ( ) ( )3

2 2 2 2 * i- * * 3- i- - 3*= + = − − −

⇒ 2 2 2 2!x* 3- - 3*

* -= − = −

( )2 2

!x

* -2

* -

+⇒ = −

+.

&. ( )22 22 1 1− = + ( ) ( )

4 22 2 1 1 2 1⇒ − − = + +

2 2 2 + +⇒ + + = ⇒ + =⇒ R () = + ⇒ lies on the i"a#inar! axis.

'. A.A =1 + ++ 1 +

+ + 1

=

.

$. A = ⇒ A(1+ ) = 1+

1 1 1 4 2 2 1+ + & 1 + +

2 1 3 & + + 1+ & 1+ + 1 +

1 1 1 1 2 3 + + & + + 1

− − α ⇒ − − α = α − = − + α

if &α = .

.

n n 1 n 2

n 3 n 4 n &

n ' n $ n

lo#a lo#a lo#a

lo#a lo#a lo#a

lo#a lo#a lo#a

+ +

+ + ++ + +

<3 → <3 / <2, <2 → <3 / <1

=

n

n 3

n '

lo#a lo#r lo#r

lo#a lo#r lo#r

lo#a lo#r lo#r

+

+

= + (here r is a co""on ratio).

. Let nu"bers be a, b a b 1, ab 4 ab 1'⇒ + = = ⇒ = , a an b are roots of the

e-uation


13

A:::0%A%:R50-13



2x 1x 1' +⇒ − + = .

1+. (3)

( ) ( ) ( )2

1 * * 1 * 1 * +− + − + − = (since (1 / *) is a root of the e-uation x2 B *x B (1 / *) = +)

( ) ( )1 * 1 * * 1 +⇒ − − + + =

( )2 1 * +⇒ − = ⇒ (1 / *) = + ⇒ * = 1

su" of root is *α + β = − an *rouct 1 * +αβ = − = (here β = 1 / * = +)

+ 1 1⇒ α + = − ⇒ α = − ⇒ Roots are +, /1

11. ( ) ( ) 25 C 1 3 & ........ 2C 1 3 C= + + + + − = +5(C B 1)=1 B 3 B & B............. B (2C / 1) B (2C B 1)

= ( )2 23 C 2C 1 C 2C 4+ + + = + + Efro" 5(C) = 23 C+ F

= 3 B (C2 B 2C B 1) = 3 B (C B 1)2 = 5 (C B 1). Althou#h 5 (C) in itself is not true but it consiere true ill ala!s i"*l! toars 5 (C B 1).

12. 5ince in half the arran#e"ent A ill be before : an other half : ill be before A.

8ence total nu"ber of a!s ='@

2 = 3'+.

13. ;u"ber of balls = nu"ber of boxes = 38ence nu"ber of a!s = $<2 = 21.

14. 5ince 4 is one of the root of x2 B *x B 12 = + ⇒ 1' B 4* B 12 = + ⇒ * = /$

an e-uation x2 B *x B - = + has e-ual roots

⇒ 9 = 4 / 4- = + ⇒ - =4

4 .

1&. <oefficient of ile ter" in ( )4 4 2

3 21 x t <+ α = = × α

<oefficient of ile ter" in ( ) ( )' 3'

4 31 x t <− α = = −α

4 2 ' 32 3< < .α = − α

3' 2+

1+

−⇒ − = α ⇒ α =

1'. <oefficient of xn in (1 B x)(1 / x)n = (1 B x)(n<+ /n<1x B SS.. B (/1)n /1 n<n / 1 x

n / 1 B (/1)n n<n

xn)

= (/1)n n<n B (/1)n /1 n<n / 1( ) ( )

n1 1 n= − −

.

1$. ( )n n n

n nr n r n n n

r + r + r +r n r r

r n r n r t < <

< < < −

= = =−

− −= = = =∑ ∑ ∑ Q

n n

n n nr + r +r r

r n r n2t

< <= =

+ −= =∑ ∑

n

n nnr + r

n 1 nt 5

2 2<=

⇒ = =∑ n

n

t n

5 2⇒ =

1. ( )"

1T a " 1

n= = + − .....(1)

an ( )n

1T a n 1

"= = + − .....(2)


14

A:::0%A%:R50-14



fro" (1) an (2) e #et1 1

a , "n "n

= =

8ence a / = +

1. f n is o then (n / 1) is een ⇒ su" of o ter"s ( ) ( )2 2

2n 1 n n n 1

n

2 2

− += + = .

2+.2 4 'e e

12 2@ 4@ '@

α −α+ α α α= + + + + SS..

2 4 'e e1 .......

2 2@ 4@ '@

α −α+ α α α− = + + +

*ut α = 1, e #et

( )2

e 1 1 1 1

2e 2@ 4@ '@

−= + + + SSS..

21. sin α B sin β = 21'&

− an cos α B cos β = 2$'&

− .

5-uarin# an ain#, e #et

2 B 2 cos (α / β) = 2

11$+

('&)

⇒ 2

cos2 13+

α − β = ÷

⇒3

cos2 13+

α − β − = ÷

3

2 2 2

π α − β π < < ÷ Q .

22. 2 2 2 2 2 2 2 2u a cos b sin a sin b cos= θ + θ + θ + θ

=

2 2 2 2 2 2 2 2a b a b a b b a

cos2 cos22 2 2 2

+ − + −+ θ + + θ

⇒

2 22 2 2 2

2 2 2 2a b a bu a b 2 cos 2

2 2

+ −= + + − θ ÷ ÷

"in alue of 2 2 2u a b 2ab= + +

"ax alue of ( )2 2 2u 2 a b= +

⇒ ( )22 2

"ax "inu u a b− = − .

23. reatest sie is 1 sin cos+ α α , b! a**l!in# cos rule e #et #reatest an#le = 12+ο.

24. tan3+° =h

4+ b+⇒ 3 h 4+ b= + S..(1)

tan'+° = hIb ⇒ h = 3 b S.(2)

⇒ b = 2+ "

2&. 2 sin x 3 cos x 2− ≤ − ≤ ⇒ 1 sin x 3 cos x 1 3− ≤ − + ≤⇒ ran#e of f(x) is E/1, 3F.


1&

A:::0%A%:R50-15



8ence 5 is E/1, 3F.

2'. f ! = f (x) is s!""etric about the line x = 2 then f(2 B x) = f(2 / x).

2$. 2 x +− > an 1 x 3 1− ≤ − ≤ ⇒ x E2, 3)∈

2. 2

2

1 a b2x 2x

a b x x 2ax x2 2x x

a b a bli" 1 li" 1 e a 1, b R

x xx x

÷ ÷ × × + ÷ ÷ + ÷

→∞ →∞

+ + = + + = ⇒ = ∈ ÷ ÷

2.x

4

1 tan x 1 tan x 1f(x) li"

4x 4x 2π→

− −= ⇒ = −

− π − π

3+.! . .. ..! e! ex e

+ ∞++= ⇒ x = ! xe +

⇒ lnx / x = !

⇒

! 1 1 x1

x x x

−= − = .

31. An! *oint be 2

t , t2

÷

ifferentiatin# !2 = 1x

⇒! 1 1

2 (#ien) tx ! t 2

= = = ⇒ = .

⇒ %oint is

, 2

÷

32. f ″ (x) = '(x / 1) ⇒ f ′ (x) = 3(x / 1)2 B c

an f ′ (2) = 3 ⇒ c = +⇒ f (x) = (x / 1)3 B C an f (2) = 1 ⇒ C = +

⇒ f (x) = (x / 1)3.

33. :li"inatin# θ, e #et (x / a)2 B !2 = a2.

8ence nor"al ala!s *ass throu#h (a, +).

34. Let f ′(x) = 2ax bx c+ + ⇒ f(x) =3 2ax bx

cx 3 2

+ + +

⇒ ( )3 21f(x) 2ax 3bx 'cx '

'

= + + + , ;o f(1) = f(+) = , then accorin# to Rolle>s theore"

⇒ f ′(x) = 2ax bx c ++ + = has at least one root in (+, 1)

3&.

r n

n

nr 1

1li" e

n→∞=

∑ =

1

x

+

e x (e 1)= −∫

3'. %ut x / α = t

⇒ sin( t)

t sin cot tt cos tsint

α += α + α∫ ∫ ∫

= ( )cos x sin ln sin t cα − α + α +


1'

A:::0%A%:R50-16



A = cos , sinα = α

3$.x

cos x sin x−∫ 1 1

x2

cos x4

=π + ÷

∫ 1

sec x x42

π = + ÷ ∫ =

1 x 3lo# tan <

2 2

π + + ÷

3. ( ) ( ) ( )1 1 3

2 2 2

2 1 1

x 1 x 1 x x x 1 x

−

− −

− + − + −∫ ∫ ∫ =

1 1 33 3 3

2 1 1

x x xx x x

3 3 3

−

− −

− + − + − =2

3.

3. ( )

( )

22

2+

sin x cos xx

sin x cos x

π

+

+∫ = ( )

2

+

sinx cos x x

π

+∫ = 2+

cos x sin x

π

− + = 2.

4+. Let =+

xf(sinx)x

π

∫ =

+ +

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

π π

π − = π −

∫ ∫ (since f (2a / x) = f (x))

⇒ = πI 2

+

f(sinx)x

π

∫ ⇒ A = π.

41. f(0a) B f(a) = 1

1 =

f(a)

f( a)

x#{x(1 x)}x

−

−∫ = ( )f(a)

f ( a)

1 x #{x(1 x)}x

−

− −∫ ( ) ( )b b

a a

f x x f a b x x

= + − ÷ ÷

∫ ∫ Q

21 =

f(a)

f ( a)

#{x(1 x)}x

−

−∫ = 2 ⇒ 2 I 1 = 2.

42. Area =

2 3

1 2

(2 x)x (x 2)x− + −∫ ∫ = 1.

43. 2x B 2!!′ 0 2a!′ = +

a =x !!

!

′+′ (eli"inatin# a)

⇒ (x2 / !2)!′ = 2x!.

4&. ! x B x ! B x2! ! = +.

2 2

(x!) 1! +

!x !+ = ⇒

1lo# ! <

x!− + = .

4&. f < be (h, C) then centroi is (hI3, (C / 2)I3) it lies on 2x B 3! = 1.


1$

A:::0%A%:R50-17



⇒ locus is 2x B 3! = .

4'.!x

1a b

+ = here a B b = 01 an4 3

1a b

+ =

⇒ a = 2, b = 03 or a = 02, b = 1.

8ence ! !x x1 an 12 3 2 1− = + =− .

4$. "1 B "2 =2c

$− an "1 "2 =

1

$−

"1 B "2 = 4"1"2 (#ien)

⇒ c = 2.

4. "1 B "2 =1

4c, "1"2 =

'

4c an "1 =

3

4− .

8ence c = 03.

4. Let the circle be x2 B !2 B 2#x B 2f! B c = + ⇒ c = 4 an it *asses throu#h (a, b)

⇒ a2 B b2 B 2#a B 2fb B 4 = +.

8ence locus of the centre is 2ax B 2b! / (a2 B b2 B 4) = +.

&+. Let the other en of ia"eter is (h, C) then e-uation of circle is(x / h)(x / *) B (! / C)(! / -) = +%ut ! = +, since x0axis touches the circle

⇒ x2 / (h B *)x B (h* B C-) = + ⇒ (h B *)2 = 4(h* B C-) (9 = +)

⇒ (x / *)2 = 4-!.

&1. ntersection of #ien lines is the centre of the circle i.e. (1, − 1)<ircu"ference = 1+π ⇒ raius r = &

⇒ e-uation of circle is x2 B !2 − 2x B 2! − 23 = +.

&2. %oints of intersection of line ! = x ith x2 B !2 − 2x = + are (+, +) an (1, 1)

hence e-uation of circle hain# en *oints of ia"eter (+, +) an (1, 1) is

x2 B !2 − x − ! = +.

&3. %oints of intersection of #ien *arabolas are (+, +) an (4a, 4a)

⇒ e-uation of line *assin# throu#h these *oints is ! = x

Qn co"*arin# this line ith the #ien line 2bx B 3c! B 4 = +, e #et

= + an 2b B 3c = + ⇒ (2b B 3c)2

B 2

= +.

&4. :-uation of irectrix is x = aIe = 4 ⇒ a = 2

b2 = a2 (1 − e2) ⇒ b2 = 3

8ence e-uation of elli*se is 3x2 B 4!2 = 12.

&&. l = cos θ, " = cos θ, n = cos β

cos2 θ B cos2 θ B cos2 β = 1 ⇒ 2 cos2 θ = sin2 β = 3 sin2 θ (#ien)

cos2 θ = 3I&.

&'. ien *lanes are


1

A:::0%A%:R50-18



2x B ! B 2 − = +, 4x B 2! B 4 B & = + ⇒ 2x B ! B 2 B &I2 = +

9istance beteen *lanes =1 2

2 2 2

H H

a b c

−

+ +=

2 2 2

H & I 2 H $

22 1 2

− −=

+ +.

&$. An! *oint on the line 1

! ax t (sa!)

1 1 1

+= = = is (t1, t1 / a, t1) an an! *oint on the line

( )2

!x a t sa!

2 1 1

+= = = is (2t2 / a, t2, t2).

;o irection cosine of the lines intersectin# the aboe lines is *ro*ortional to(2t2 / a / t1, t2 / t1 B a, t2 / t1).8ence 2t2 / a / t1 = 2C , t2 / t1 B a = C an t2 / t1 = 2CQn solin# these, e #et t1 = 3a , t2 = a.8ence *oints are (3a, 2a, 3a) an (a, a, a).

&. ien lines! 3 ! 1x 1 1 x 2

s an t1 1I 2 1 1

+ −− − −= = = = = =

−λ λ − are co*lanar then *lan

*assin# throu#h these lines has nor"al *er*enicular to these lines

⇒ a 0 bλ B cλ = + ana

b c +2

+ − = (here a, b, c are irection ratios of the nor"al to

the *lan)

Qn solin#, e #et λ = 02.

&. Re-uire *lane is 51 / 52 = +here 51 = x2 B !2 B 2 B $x / 2! / / 13 = + an52 = x2 B !2 B 2 / 3x B 3! B 4 / = +

⇒ 2x / ! / = 1.

'+. ( ) 1a 2b t c+ =r r

S.(1)

an 2b 3c t a+ = S.(2)

(1) / 2×(2) ⇒ ( ) ( )2 1a 1 2t c t ' ++ + − − = ⇒ 1B 2t2 = + ⇒ t2 = 01I2 t1 = 0'.

5ince a an c are non0collinear.

%uttin# the alue of t1 an t2 in (1) an (2), e #eta 2b 'c ++ + = .

'1. ?orC one b! the forces 1 2 1 2O an O is (O O ) + × , here is is*lace"ent

Accorin# to -uestion 1 2O O+ = J J J J J J J J J(4i K 3C) (3i K C) $i 2K 4C+ − + + − = + −

an J J J J J J J J J (&i 4 K C) (i 2 K 3C) 4i 2K 2C= + + − + + = + − . 8ence 1 2(O O ) + × is 4+.

'3. <onition for #ien three ectors to be co*lanar is

1 2 3

+ 4

+ + 2 1

λλ −

= + ⇒ λ = +, 1I2.

8ence #ien ectors ill be non co*lanar for all real alues of λ exce*t +, 1I2.

'3. %roKection of alon# u an alon# u is u

H u H

× an

u

H u H

×res*ectiel!


1

A:::0%A%:R50-19



Accorin# to -uestion u u

H u H H u H

× ×= ⇒ u u× = × . an +× =

2 2 2 2H u H H u H H H H H 2u 2u 2 − + = + + − × + × − × = 14 ⇒H u H 14− + = .

'4.

( )

1a b c b c a

3× × =

r r r

⇒

( ) ( )

1a c b b c a b c a

3× − × =

r r r r r r

⇒ ( ) ( )1

a c b b c b c a3

× = + × ÷

r r r r ⇒ a c +× = an ( )

1b c b c +

3+ × =

⇒ 1

b c cos +3

+ θ = ÷

⇒ cosθ = /1I3 ⇒ sinθ =2 2

3.

'&. oe can be co"*ute fro" histo#ra" an "eian is e*enent on the scale.8ence state"ent (a) an (b) are correct.

''. ix a for i 1, 2, .... ,n= = an ix a for i n, ...., 2n= − =

5.9. = ( )2n 2

i

i 1

1 x x2n =

−∑ ⇒ 2 =2n 2

i

i 1

1 x2n =

∑ 2n

i

i 1

5ince x +=

= ÷

∑ ⇒ 212 2na

2n= × ⇒ a 2=

'$. 1: D eent enotin# that A s*eaCs truth

2: D eent enotin# that s*eaCs truth

%robabilit! that both contraicts each other = ( ) ( )11 2 2% : : % : :∩ + ∩ =4 1 1 3 $

& 4 & 4 2+× + × =

'. ( )%(: O) %(:) %(O) % : O∪ = + − ∩ = +.'2 B +.&+ / +.3& = +.$$

'. ien that n * = 4, n * - = 2 ⇒ - = 1I2 ⇒ * = 1I2 , n = ⇒ *(x = 2) =

2 '

2

1 1 2<

2 2 2&'

= ÷ ÷

$+. % B P = 4, %2 B P2 = ⇒ % =1 1

2 2 ; an P 2 2 ;2 2

+ = − ÷ ÷

.

$1. O . 3 sin θ =

O . 4 cos θ = 1'

⇒ O = &.

$2. ! La"i>s theore"

% D P D R = A <

sin + D sin + D sin +2 2 2

° + ° + ° + ÷ ÷ ÷

⇒ A <

cos D cos D cos2 2 2

.


2+

A:::0%A%:R50-20



$3. Ti"e T1 fro" A to =12

4 = 3 hrs.

T2 fro" to < =&

& = 1 hrs.

Total ti"e = 4 hrs.

Aera#e s*ee =1$

4 C"I hr.

Resultant aera#e elocit! =13

4 C"Ihr.

$4. <o"*onent alon# Q = ( )

1sin3+

14 ' 2sin(4& 3+ )

°= −

° + ° "Is.

$&. t1 =

2usin

#

α

, t2 =

2usin

#

β

here α B β = +

+

∴2

2 21 2 2

4ut t

#+ = .

FIITJEE Ltd ICESH S i Vih (N H Kh B T )N D lhi 16 Ph 26865182 26965626 26854102 26515949 21

A:::0%A%:R50-21

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