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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
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$. 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
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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$. f
n
n nr + r
15
<=
= ∑ an
n
n nr + r
r t
<=
= ∑ , thenn
n
t
5 is e-ual to
(1) 1n2
(2) 1 n 12
−
(3) n / 1 (4)2n 1
2
−
1. Let r T be the rth ter" of an A.%. hose first ter" is a an co""on ifference is . f for so"e
*ositie inte#ers ", n, " ≠ n, " n
1 1T an T
n "= = , then a / e-uals
(1) + (2) 1
(3)1
"n
(4)1 1
" n
+
1. The su" of the first n ter"s of the series 2 2 2 2 2 21 2 2 3 2 4 & 2 ' ...+ × + + × + + × + is( )
2n n 1
2
+
hen n is een. ?hen n is o the su" is
(1) ( )3n n 1
2
+(2)
( )2n n 1
2
+
(3) ( )
2n n 1
4
+(4)
( )2
n n 1
2
+
2+. The su" of series1 1 1
...2@ 4@ '@
+ + + is
(1)( )2e 1
2
−(2)
( )2
e 1
2e
−
(3)( )2e 1
2e
−(4)
( )2e 2
e
−
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21. Let α, β be such that π α 0 β 3π. f sinα B sinβ =21
'&− an cosα B cosβ =
2$
'&− , 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$. The o"ain of the function( )1
2
sin x 3f(x)
x
− −=
−
is
(1) E2, 3F (2) E2, 3)(3) E1, 2F (4) E1, 2)
2. f
2x
2
2x
a bli" 1 e
x x→∞
+ + = ÷
, then the alues of a an b, are
(1) a R, b R∈ ∈ (2) a = 1, b R∈
(3)a R, b 2∈ = (4) a = 1 an b = 2
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2. Let1 tan x
f(x) , x , x +,4x 4 2
− π π = ≠ ∈ − π . f f(x) is continuous in +, , then f
2 4
π π ÷
is
(1) 1 (2)1
2
(3) 1
2− (4) 01
3+. f ! ... to! ex e
+ ∞+= , x G +, then!
x is
(1)x
1 x+(2)
1
x
(3)1 x
x
−(4)
1 x
x
+
31. A *oint on the *arabola 2! 1x= at hich the orinate increases at tice the rate of the
abscissa is(1) (2, 4) (2) (2, 04)
(3)
, 2
− ÷
(4)
, 2
÷
32. A function ! = f(x) has a secon orer eriatie f ″(x) = '(x / 1). f its #ra*h *asses throu#h
the *oint (2, 1) an at that *oint the tan#ent to the #ra*h is ! = 3x / &, then the function is
(1) ( )2
x 1− (2) ( )3
x 1−
(3) ( )3
x 1+ (4) ( )2
x 1+
33. The nor"al to the cure x = a(1 B cosθ), ! = asinθ at θ> ala!s *asses throu#h the fixe*oint(1) (a, +) (2) (+, a)(3) (+, +) (4) (a, a)
34. f 2a B 3b B 'c =+, then at least one root of the e-uation 2ax bx c ++ + = lies in the interal
(1) (+, 1) (2) (1, 2)(3) (2, 3) (4) (1, 3)
3&.
r n
n
nr 1
1li" e
n→∞=
∑ is
(1) e (2) e / 1(3) 1 / e (4) e B 1
3'. f sinx
x Ax lo#sin(x ) <sin(x )
= + − α +− α∫ , then alue of (A, ) is
(1) (sinα, cosα) (2) (cosα, sinα)
(3) (0 sinα, cosα) (4) (0 cosα, sinα)
3$.x
cos x sin x−∫ is e-ual to
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(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!
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(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+ +.+ +.+$ +.+&
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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
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$&. 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
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FIITJEE AIEEE 2004 (MATHEMATICS)
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
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FIITJEE AIEEE 2004 (MATHEMATICS)
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
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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)
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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.
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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 +
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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.
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⇒ 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
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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!
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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
.
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2+
A:::0%A%:R50-20
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$3. Ti"e T1 fro" A to =12
4 = 3 hrs.
T2 fro" to < =&
& = 1 hrs.
Total ti"e = 4 hrs.
Aera#e s*ee =1$
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
#+ = .
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A:::0%A%:R50-21