JEE Sample Paper 5

21
7/23/2019 JEE Sample Paper 5 http://slidepdf.com/reader/full/jee-sample-paper-5 1/21

Transcript of JEE Sample Paper 5

Page 1: JEE Sample Paper 5

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Page 5: JEE Sample Paper 5

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Page 6: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 6/21

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i>

do

ubled, k e ~ p

co n cen

rrau o n

of B a

rul C oon<tant

, rh

t M e

of IC

"-<tion becom

t< dnubi

e.

(iiJ i f

con,..,otr

ation ul

 JJ hah

'Cli,  '<eep

ing  

co r.cen

tration

ul A

nd

C

 co rntnnt,

 t he 

ra t

t

of

 1 ea ctiun

re •nain '

unaf

k' ed.

(

1ti)

if

co nccntra

<ton o

f

C

i> made 1.

5 time

,, d e

ra te o

f re action

 bc,·omc<

 2.25 tim

e,_  

T he orde

r of reat·n

on "

(b) 2 .5

(d

) 3 5 

A

hy drogen

 

g

a:, ck

<l <ude

hn> poL

erttial of

 

- 0.118

 Vwhc

uH, g

as

is

a

t 2 98 K an

d l 

a[JU.

in ]-;CI solu

tion.

ll•c

pl l

of H

CI solu:io

n i<

 

a)2

 

b) l

(

c ) 7 

(d ) 2  7

 

2 Jf log

  ~ l o g

nd log

g I <O

in

AP

'

_Sc

, . ~ - - ~

wh

ere

a

 

c o

re m  GP,

then a 1

, c are

th e 

length

; of sk cs

 o f

ta l

 

an

 ;,' "' '"-

los <ria.ngle

{b ) 

an

equilattrai

t f l a l l ~ l

(c)

a ><a

lene n iangle

 

(d)

N

o11e

clth

e abo)Ve

 

1

1.

 

I

Page 7: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 7/21

C

e

n:

r<

'

o

dt

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e '

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(4

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tw

o

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qr

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;,

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,,

;, o

l i

n

  '

 

4.

 

I f

 l

b

m

  -8

1

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l

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th

e

o

f

 

1 he

 

ci

rc

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h

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nn

g

 U

m

_v

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0 i

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T 2

y

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en

 

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l

 

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I

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p

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o

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r

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lit

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en

 

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rj

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t

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n-

IQ

 

l

J

 

L

 

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c)

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n

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l

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.,,

Page 8: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 8/21

[o) d

2

( x ' -

y')

(c)

d'Cx+y)

CbJ

a'

c ~ ·

_ _

.v l

(d) <one of

Ll•e>e

.;.O;re< iol>S : ~ ' ~ ' ' > i o . 1Q ' o r < A « n t f ; , : ~ '

·

~ ' Uldl of (lu >. ~ ' ' ' · ''' ' tri'-: '

W t t t 8 ~ n < n < t ="""' 1 tA<IrniorU,

o • t ~

10

m,.,f f i)

Fool. oft/,e,<e

q1fe.<ti 'al>ni= /ouT

~ l t < . , t l n

cJ.m'is. '"1"'""'

of

•-him

4 W

t w t ~ < ¥ · \,"

o,;.,_ to

;dectdl; '""""W,t<t"in " <e = f p s ( ~ ) , (bJ, (0 .,;,JJ<l);m,.

-th<sivmMOw .

---

no

.

t a ' ~ ' ' " <nre, s t a = t 14 . 'sta. ' R' l

.

"'""

- """"""'""''"''

,jt"""'""'

.(bl """' '""' J.is ~ Stan,'""'" ,fi1o<.,

W ~ ' n r f i : f o ~ ~ S W < m m r i i L i m r ~ , - _- _·

[,j]

S > o t < m M l 1 ~ - ~ l u ' i O n ] o r . s . & t . ~ i · f _

- ~ , -'1) '

6 State<nentl : tan-

1

(

4

I :an · [:7

4

St.tement : Fot x

>

0, y 0

tan· [ )+

<an- [·y-:<J

Y

) +X ,

4

17. Statemffit 1 The hne 2.x+y > 6 ~ 0 i5

pcrp€Ildiculat Lo the me -

2y

+ 5

=

0 <1ml

S€oond line pa«e< tbruugh (1,

3)_

Statement

II

:Product of the slop<> uf the lines

"'ocu,,)

w-1 .

18. St...teme.nt

' 'dx

=

0 0 ~ .

' e • ~ x d x - i . .

' '

.

. .

Stntementll:j f { : t ) ~ n j j (x ) jx ,oEiond

. '

j [ a + x ) ~ J ( x l

'

'

tatement : l f l '(x and

( : . - J ~ " '

polynondals,

c ,.

P(x)

c_ , c , .

tuen

.': ':.

Q(x) 11 a'""'"

non->x,oco

nmnuers,

u

degree of

P(x)

and Q(x) are equal

20. Theequ"Lion

of

the hyperl>ola possing through

the point ll,- I ""d having <S)'ln:''"''"

x+

2y 1- 3 ~ 0 a n d 3x-l 4y + S ~ O i s

(a) + B ; c ' + J O : r y + l ~ x ~ 2 2 y ~ n

Cbl 3.>:' + sy' T JOzy

+'4x

+ 22y + 7

'

rl

s x · ~ s y

., > y + l 4 x + 2 2 _ y + 7 ~ o

(d)

3x

2 I

H e ' ' ~

10:ry

+

14.< F

22y

7

=

0

21. GP coruim nfan evennumber uftenn,_ If the

sum uf all die "'rn"

lS

5 umes the

mm

of tho

terms ocrnpying odd place.<, tho common rario

Will

i>e

equal to

(a )2

(b )3

( c ) ~ ( J )S

22. The equallon of the plane throu3h the point

{2, -1, - : f ) and parallel w the line<

x- 1

y+2

• x

y - 1

z - 2

3 2 - - - - 1 o n d 2 ~ - - - - : : : : f ~ - 2

(a)

8 - < + 1 4 y . l 3 z + 3 7 ~ o

·

(b)

8 x - 1 4 y + 1 3 z + 3 7 ~ 0

(c) 8 x + l 4 y - 1 ~ • + 3 7 ~ D

(d) None of \he abuve

The

AM of ''"'c

0

, ,_,C

1

•'c,,

'

(a)

'

'

"'

'"

"-

,_,

Tho number of real ..-,)utions of

1+I< -11=<"

to"-

2lis

(a)

U

(cl 2

(h)

1

(d) 4

...... , '"''c,

The three roots of the

eqll<Uion

:1= G

7 6

'"

a) -9,2,7

(bJ9, z,7

(c)

9,2,-7

(d) Koucufthese

l f and r deno>e a tauto lozy arui

comradictwn

respectlve;y, then for any statemem

p

p

A

r

i<

equal to

(0) '

(bi t

(c) r

(d) None

of

the>e

The

gceatest value

e>f

x

2

y

3

,

when 3x

+

4)

5,

a;

'

c) 5 ( d ~ None of these

lflm (

2

"

+ ; 1

- 2, then the lorus ofthe point

' '

representing < 111 th•· complex plane;,

(a)

a

mcle

(b) a straight line

(c)

rl

pHrabola

(d) None of Lhe above

Page 9: JEE Sample Paper 5

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(a

)

7

(<:)

0 ' 

(d

) No

ne o 

th

 

31 . A

  co

ne of

 

m

ax im

um  vo

lmne

1> lio

>n ;h,J

  111  o

W J \

sp il•l

e, th

en ra ti

o

ofh

eipt

 o

f ho

wn

10

 

d

iaaet

er

ot

th e >-

phere

;,

  ~  

h

 2

'3

'' '4

 

el

l

c

,

4

3

2 . T

he ore

a of  o

ne tun l

liueor

 

tl  

iang

le  f

01m ed

  by

th e CU

P;<>) = 

,i n

 

X

= O(

}.U and

N

IX i' i:;

(a) (2

+,·'2J

<qun

it

(b)

2

,'

2J sq

 unit

(c) (

/7.

 

-

2

)

sq u

ni t 

(

d)

N

one

of Lhe

 abo

ve

••

P

HYSIC

S N

D CH

EMIS

TRY

) .

(c)

'

(

0)

(c) 

••

'

'

(

C)

n

.

'

 

·

 

(

c)

'

'

'

(c

)

(;)  

,,,

(•)

· 

·

)c

)

' '

·

'

' '

.

"'

~  

' '

...

  (cj

«

 

(0

}

· 

'

'

.

.

 

'

(o)

· 

'

 

... 

'

'

...

'

'

· 

.

..

'

... 

,,,

• 

M

 THE

M   TI

CS

).

'

:c)

 

(b)

'

'

'

·

·

'c '

 

·

'

'

'

 

...

'

·

,

,,

' ·

 

(

d)

 

(a)

·

 

'

· 

'

' ·

( b ~

(<)

••

(a

)

... 

'

••

'

 

·

 

·

· 

••

·

vc

cL O«

.:i i , j

,k a r

.d b

= i - j

+ : lk

o n c

l

i n ~

.m  

d

tu lc

;mg

k

w

ill  o

he

ve<J.

-.o ,_

"

 

1

, ,

,

(a)

 

-

~ 4

i - j

- 3 k

)

J

26 

(b)

L

4i j

:Jii

)

.

(c)

,·2 ,

(41 - j

+ :lk) 

(

d)

of che

 ab o

ve  

Th o p

ro bab i

li ty th

at out

 of  1 0

  r < u n s

•II

 boon

 

in

Apnl

, at l eas

t tw

o ], . , , . ,

th

e same 

bm

hdaf, is 

c

(o

)

(h) 1

 

--

 

(3

0)'"  

30  

go

-

  'c-

fcJ

  --

-

(

: l ld

(tl) No

ne nf

 

L

be3e

'

(a)

'

(b)

 

• 

'

 

c)

(c)

 

· 

(C)

(c)

·

:c)

,,

w

(a )

·

 

'

 

·

,,,

(b

)

·

(c)

 

...

' '

o)

~ c )

·

... 

'

'

...

'

56. 

(o ) 

(c)

(

t·)

· 

(c

(

...

(c)

'

 ·

( )

(d)

 

'

 

·>

l

••

 

c ;

·

'

(d )

(c

)

·

 

(o )

,

'  '

c)

..

. (c )

 

·

(o

)

CO

Page 10: JEE Sample Paper 5

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I

hysics

1. p-type or n-type 'emiCnnductms are unchat](€d

» the dopants are donating or acceptin); the

€'e("tn}n>, •hu>

acquiring negotive

or

;>osil1W

cl>arge_

Charge of capacir01

is

the charge

on

facing

,-urface> of the plat€> of

capatitor.

Q ~ : _ ' t /'J

)10- ( -7 ) )

'

6itC

Potemioldifferene< across the caf dtor- l2V

_,

so.

C v

l

G

\ ~ o - )

JF

O S ~ F

3.

Due

w

the proce» <J rrnognetit

hlu<thm

ma&net can

allractnon-

magncUzcd substances.

4. The needle will stoy

in

any direcUon

it

is

relea..,d_

S-

Dire(tly rom theory.

6. The rota averoge enetll)' of SHM in o ~ e time

. .

mw

2A 2

penOO

"'

0

- -

7.

Af;

r decrease•. frequency incrtd'""

•o

""'ek"s''' •k<

' ~ ' e ' -

6. l.N

parnde

is projected l 'ito spe•d

u

>a total

rirr:e of

l i ~ h c

= Jand<=2x m a ~ i m u m h e > g h t

_

;

'

= 2 x ~ = g

n rheTe

i>

no zrdv:icy then>'

=" •

T

2u

~ - - ~ 2 >

'

gravit)' is not there, it

will

ne>-er fall kck.

9. At th< locanon of

loop_

magnetlc Held ,,

i>"tpendicular to plane

of

P P ' and going inm

i•

wlri<-h

h im;ru.,ing wilb lime d5 current i>

increasm.5. Therefore. emf

will be :nduced in

tbe loop m such a way

;o

that induced current

will

pru<lua: mogneti< l idtl ""oudo il ur'J-00'''

Lhe ong;,.,) mognet1c field m aroordance ~ i t h

Le:u s law_ ·n.u,_ currem i• induced in the loop

in antidO<·kwise du:ettion.

10 rn.><e%

;, Jsobm1c su.'JW p W

~ 1 . 0 1 X JO'

0671

-1 X

lO-o

=

168.6'

J

IIQ-Ix540x4.2J

2268 J

AL' .l.Q - AIV

=

2099 33

11 *·so tho cntw

;s

rcctaJlgular

hyporl>ola.

12. Fr<UHJ(: <Y

heaoJ by oh<crvcr (8'-"' 1onaoy) will

he

' 'mllm

if

<Olltce

mm'lng

mwarch; rlle

ob<e;Ver, 'hidl is the co"" when tile 'our<:eis

otR

Frf'l uency heard

by

observer (sco<10nary) will

he mioimuro, il;oun-<; ;, moving Hwoy fcnm the

o b ~ e t v e c ,

wh1ch

L<

Lhc

oa<o

when

Lhc wurcc

i

oLA

--·o

For

point C. JreqLtency

llean:\ is origmal

fr ' ue;rcy.

I - ~ M R '

l• . l i=_;Mil

2

0

> :le<.'JT_l

0

~ I

Ia

x

2 > T

Force

between two charge particle is

iodepenclem vf presence

of >n;-

orher charge,

(superposition plindple)

' '

- K 4 R ~ 2 K 4 . R

4

KA

f R

1

R,

I

R , , ~ , R +R

+R,_

~ 1 2 K A

' .

'

.

[e> rods 1 ond 2

' in

pa illeland equivalent

is

1n

' ' 'with

:<J

O<k

where

s

the

lSt:mce rraver>e<l

y

Page 11: JEE Sample Paper 5

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Tm•g formed by roncove lens is"'""" fur all

pos lions

of

ob)e<tre, llllag• Oyccme>lve lens lies

~ ' ' ' ' '

the " " ' len.les_

For both

the

im"l,'tS

to

col lcide, image bv mnvex lens should also lie

in betweffi <be ;,_o

I011<e<

the

rwa

lens or

ima ;e c-unvc< lens should also be >·irtual.

\

I

j_j

1_. _<

20-< ._ . . I

For convex len>

'

,

For oo:ncave lens

'

20 v

'

'20 -x )

' '

~ a l v i n ~

Llus

equorion, we get

20(, '3-l j

x

..

em

1d ' = 2 0 ( ~ ' 3 - l ) c - m

M a ~ u i l i c o l i o n by

ronvex em

'

W

...

i)

... i i)

v

1

v v 2 0 ( > ~ 3 - l )

-./3

m =;;; =X= X zo(./3-l)t 13 =

Magruficanon Dy roncove lens

v

2

-{20_-,·) 20-v

" ' ·--

'

- (20-x) 20 x

- ~ ' ~ ' ~ '

= 2-/3-3)

2 0 - ? ~ U ~ I )

,-'3

19. By cou>€rvation of momentum,

the momenhlm

of the block bullet sy<tem JUst after the

interaction

is

p

r;w,

block-bullet

>)>leou

on the

able top.

From work-•ne.rgy

theorem.

t K

=WI

0 - K = - ~ . ( m + M J g <

'=0.37

m

20. 'Jhe work dor_r by gravity is the "ark done. as

if all the rna"

were

con,entrared

at

the

<enuc

of mass. 1he work <H'Ce<sary to lift tbe o b _ i e c ~

am

be thol'ghi ol "

the work done

>gaul't

gravitv ond

is

just

W

=

m i[h

where

"

is

the

heigll: through which tbe centre of

m =

is

rui>ed.

W

= (180

kg]

(9.8 rn/.<

1

) I 7

m)

= 3.0 kJ

C'urrent throLOgh harrery

having em E

JS

I

A,

while

through

12 V bmery is

'hi> can be ('"k"la.-.d a•

follows.

Potential difteren<'• acm«

12VbaucryJS

f l U - 1 ~ 2

=> '=3A

E,1 l

w h ~ c c i current flowlnz through tmonob

c u n w u i u ~ l2V batle<:)'·

From Km:hhoff's circ"Uit I"" al L C\lrrent

rhrough rechargeable

"""''l'

islA_

Net power

('<JOilHnCct

= V X = 6 x l= 6 W

22. The descendirlz pari ofrhe mpe is in tee fall; It

las speed v =,.'2p;y "'the'""""' alit" point>

~ ~

4

n c ~ T r ? l i e F ~ ~

Page 12: JEE Sample Paper 5

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 ·

lw>c de>cemled a d i s r a ~ c c y. lhe length of the

rope which l•nd<

OlJ

the

tob e during

""

intf >V•I

dt

tollowing

tim inmnt lS

vdt. The

increment of momentum imparted to <ho table

hy tlri> length

in rommg

m

teSt

is m(vdt)v.

Thuo, the rate or

whrcO

momentum is

"""''enctl

w

the table;,

o,p-m>

2

- (Zmy)g

and

thi.1

1S t ~ e

force arising from •topping the

downward '>II of the rope. Since, a lengtf, of

rope y, of

the WeJght(n()' g

olro,dy

he< nn the

mbletop, lhc·

tut.J

force on the ra\llecop

Is

(2my Jg + {m_v)g ~ ( C > t r y ) g , or the weight

of"

length :\y of rope.

So, J{-3

The charge on

bmh

the cap>dtors C

1

and c,

would be same " hoth m• l'-"lllected in series,

tllen equivalent circuit

can be d f d w ~ &s and

q = C E l - < r/>X'l

w h e ~ e C -

_5_,C,

c,

~ c ,

Find acccleranan

2mc. = mg

sin

53" ·

W &

cos

53°

- mg sin 37' -IJJ"g 0<><37°

' 'hen

u<e v ="' +

2as, v= G, u= 1

m/s

The """nona

and

tramlauo,al

Krl

of the ball

at

the bottom wrll i>e c.hdnged to gravirational

potendal Oll"'lfi, whrn <he ophere stops. We

therefore ;nite

I

M>

2

Jw )

+2

=(Mgh)

 

J

'

2Mr· ;·

For

a

<olirl sphere, 1= -,-- Al<o, lO = r Then

ahovc cquatlon oecome<

_ _M-t'+

l ( ~ M r ' 1 i . " ) ' =M

0

h

2

2 \ 5

1

r o

I 2 1 ,

or

: '' -

5

v

=(9.8)0

U<.ing ~ 2 0 m ; ~ gi,•es h

=2&6

rn

Note' rh, ' """ '"0 ' ' - " " dc•ceoc ""'" ' " ' "'"" ,.

' '

b ll '

"i""

-'" ""

'"-

Rote at which ene'l)y is

mddent

on >phe"',

P

=

intensny

projec-tion

area

of

sphere

= 10

x

10

·' x

x (2x lo->f

This energy incident on sphere

impam

some

mornentum it, whicll. e ~ e r t > a Ioree on it.

dp

=

Pt-

p,

(P/<)LI

dt Jt

= ' _ = 4 . i 9 x l O " - ~

'

2.7. Acceleration due to moon's gtm1ty on moon';

surfa<e is gJ because

~ - '

~ . :

M,

(g

=

Gm .

6R:

R

While oce<lerannn dHe to e<mh'' grnvity on

moon's surface s d p p r o x i r n a t e l y ( ~ i ' or ~ 0 u

Thi• is because

chsLance of

rnoor_ from the

eorth's <>'ntte is appmximalcly e4ual to

00 times the TI\dius of

earth

""d

-- ,.This

,,

can

X

undc,..tood from

d:te

f.gure

Moon

eo•

/

,.' \ 60 R.

g,,

~ + / - · "

\,_.(--/ ___ _ - - ~ . ~ )

~

g1

= r f ~ ) ' whiles,

= ¥ ;

Ilelium

is

monoatomic gos.

while·

M)"Ken is

dmromic. Therefo,-.,, the h<'<lL ~ i Y e n to helium

will

b<

totally c c ~ e d i<• increa;in.<l the

translational kinetic eucr;,-y of it> rnoleruk<;

whereas the

h""t giV<"n to ox)"gen

will be uoeJ

up lD i n c r . . s m ~ the nan.slatio»al kinetic

energy of the molecule

and al"'

is 'llUCO>in):

the kine"c energy of rotlltioto oncl •ibration.

lienee,

there

will

be g:"eatcr rise

in

the

tcm """tllr<O

of heliwn

_

The

pcriOO

of Lhc liqutd ex<cuting SliM in ,

U

-tube

de>es not

dcp<nd upon the delL<oi (y nl n, e

liquid. Th""'fure, time period will be tho

<arne,

when merrill}' is filled up to the same o e i ~ h t " '

the water

in

the · - : u b < .

Now, as

the

pendulum oodllate,, it Jroz:; air

a l o n ~ with it_ Therefore, its kinetic L nCIJ, Y i l

dis<.ipated m

owreorning

visoom drag

due

to

air and

hence,

its

ampl tude goes on

docreasin&.

35_

Clear )'

the

courdmates

o are

)J, 2/

l

Page 13: JEE Sample Paper 5

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hemistnj

36.

Ferro<"<'ne

,,

o

.<andwJCh

complex comrmm<l m

which all the five

corbon

attnns of

cyclOjlentadiene

"nion

""'1m

ked

ro

the

metal

1hrnugh

p1

loonds.

37. F•cLuaL

38.

All arc hav'ng

paired electrons only

inC,, K,

<Lrnl F,.

39. After achleving

4 /

Sd

0

fu-'' c o n f i ~ u r d L i u u ,

lh<

next

electron goc1 to 5d

and

this is the taso

of

Lu

(Z

=

711 wlm·h is the last

element

uf

lonthanide serie5.

40.

Silicones

oom"ir- - - , S • -0 -S i -

bond•.

41.

The member\ of

x y g o ~ family

formMF

4

type

tenafluoride> (where M

-O,S.Se,

Tel,

in

whkh the central awm ;, .<r d hybrid ed with

"

'""'gulor tetrahednd g<'<>meory due

to

pres•ncc

of

one lane pair o el••<'"""'-

"'

42

m

~ C l - 1 c w < ~ r n ,

N£1 '

'

'

l ------- > CII

2

-CH .

""'''l"o-ee I •

~ ~ ; ' Cl

43-

NaNl-1,--------> Na• Nft,

Solo NH;

is

much

I""' mong

base.

me

f

' "

01 l

" CH

Ac<tk anhydrJde

' ClJ,OH

f

40

'

20 e n•

0

,

01-IC-+ CH----+

4

CH,

I

o C

----<.;H,

c

I

0 ~ C C

P'" '"" ' 'Yl

substituted gluCOIC COJlftrms the

p.-eseru:e of five hydroxy

group.

Tollen\

rted)(l'IIt

;hows the existence

of

aldd•;d;,_ group.

Propmol

(CH,CH,Cll,OH), c,H,O e<ll not he

1he ~ o m e r of C

1

 -i

6

0.

N ~ m b e r

of

plwroehtmm

emitted

Jepc,ds

UJ>Oll i ~ t e n s i t ) •

cf

nddc·JL( 1<d '"Lions_

112

g

contamed hy l mL_

then 200 rnLcontOlilS

~ I , \2x

200= 2: 4

go

HCJ,olulioo

Actual w<ight

o' HCI =

60'%

of

224 -

134.4

1344 -

:-lumber

o

moles = :lb'i - 1.M mol

Pmcess

is ciispL,cement

,;,nply

c u ' ~ is

dbp).,·ed

'>yZn o. bm the concrntra<ion

of so:;- with both jons - r n . ~ oott<tant

¥e

  93

Ouo

O"d•llott

number

of Fe

0

_,

- -2(0 .93molpo,cssnd 1

2chorge)

Suppose x mol- 2charge.

then 0.

93 -

.t

po;,;e' + :l charge

1-lentt, ,. x 2 •

3(0.93-

x ~ + 2

X •

0.79 Fe

0

•;

Fe

' •

0.14

, ~ , , _

U.l4

' l l _ , •

~ - '

re

.---X•><

_ ~ . v o n

0.93

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a

them

ati

cs

I

0 ''

'

' )

- ~ ~

- ~

 

. S

ince,

, ,X +

\ -

2

+

2

'

[u

3 r

- •

0

11 ' 

I

3• ' 

(< + lY

}

=

;  

' '

·

3'

0

(

-

m

} ''

3

  (x

- iy

)=3'

- '

=

3'

i

-- })

'

3 g,f3

x +

\ =

- z +

• - , -

3

:;.,. 3

x = -

2 'y

• - , -

· · 

0

1

'\UJ>k

 v io

le t com

plex

6

S. Th

e m a

in Plll

1'ose of

usin

g

~ n g b

ulb

is

 

to

as

sure tha

t pr

es.mre

 wi th

in the

 r<act

ion ve

.se\

lS sa

me .

,

th

at i

n th e

 room

.

69. S

tatem

em

(J)  

511&

& ''-' n

rdo w

ith r

-'Fecr

 to A i

s l

(ti l

StiAAOstS 

ord er wi th

resp ecr

ton

is

0.

(il

i) Sug

gests

o rder 

wi th r

esp<cr

 to ;, 2

-

Hen

ce, ra t

e law 

eKpre

,; ion <on

I

N wri t t

en

a

s

c=

klflJ

lB]

0

[C]

2

O

rder

ulre

,cMn

 

=

 

1 + 0 +

2 =

 3

·

.

)

_  --

  e

 

2

H

,

0.

,

1

=P

0 5 9 l

- 0 .11

8

v

0

+

  0

.0.'>9:

log ur :

 

-

o. n

s

v

 = -

0 .0

591 (-

log [

1-r+])

-0 .

118

~  

o 0

591 pH

H

= U

.l18 =

 

2

p

0 .0

591

)

X =

-

c Yo

,,

 

Bu

t

x

=

  J;y

k

=--L

 

,- 3

S

ince.

log I

5

.:'1.

g ' ~ ~

and

 

a r

c in

u_

o

c,

..

(i)

A l s o ~ . b care in GP.

2

(x+ Z y - 3 ~ - (2<+ y -3 )

Page 16: JEE Sample Paper 5

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b ~ o r

From

E.q,, (i)

>nd

(til,

we

get

9ac 25<

2

>

9a 25r

_(li)

"

5 ~ 5 < - ~ ~

[ lmm Eq.

(i)J

" , "

3 9/5

b+c<a.

Since, sum of two stdes of a triangle is smaller

than

the th11

d "d•.

. .

T r : i a n ~ l e

i> not dellned.

3.

Let

f(.<J=.<

2

-2 (a- l )x+(2o+H Then,

f x ) = 0 will have

both '

positive, I f

l_ Discriminant;, 0

2.<umofthere>o">O

3.j(0)>0

I .

Discrimir

.mt l a

'

4{o- l J ' -4 (21HlPO

2.

Stun

of the roots > o

> 2f_a - l l >

o

a> l

3.f(0)>0-'->(2<Ii 1)>0

'

> - 2

From

Eqs. (i), (i1)

and

(11i), W< get a ~

4-

Hcncc. the

lcosr

intc311tl

value

of"

is

4.

,.. Solutiom

for(/_ No.

4 to 5.

4. Given that,

16m' ~ B l +

1

=> l6 (1 ' -m' ) - J61 '+m+t

=" 1 6 C l ' ~ m ' J ~ ( 4 1 + 1 )

c-- - ,

4;' "+m ~ 1 4 1 + 1 1

1 ~ = 4

,·r,

m

" ' (i)

___ (ii)

- ( li i)

Since,

u:ntre

= ( ,

0)

md radius

=

4.

,",Equation of circle

is

-4 l ' + lY - o)'

=

4

2

5. Given t ~ u a t i o n s

oi

lines x

+

2,v- 3 0 and

:be+y- 3 0 "'" non-p<lr.Uel.

Bisecror

d

lines

(_:<+2y-3) +(2.\ '+y-3)

, , /5

>

(x+2y-3)&(2.\ '+y-3)

-

(i)

x+y

2 ... (il)

Sinoc, conlrc

of drclc

lies on angle bisector

of

lines.

Abu,

centre hes

on

d lute

3x-14y

-S

.. (ii )

On 'olving E.qs.

(i) and (iii), we

g<{?-,

;) and

><>h•ing

E.qs.

(10 and (iil),

~ gor (3,

- \ ) .

6. Since, R

n R

""" ; : ~ t dis,ioint, there is at least

one ~ e r e d p•ir,

i

c,

OJ

in R n R _

But(", b] , R n R '--"(",

b) =

R ru,j

{•>- I;)

o R' .

Since,Randl ' are 5)-Inmetric relatione,

we

g<t

(b. a h

R

and (b,o)ER"

and

comeq•oentiy(b, a) E R n R .

Similarly, onr other oro.,«

pcir

{<- d) e R n R , then we mtc• al.lo haVE

(d.

c)

El-:

,-,

1-:'.

Hence, R R

15

symrnen<£ relation.

7. >low, >inu + s i n ~ - si•, y-sin

(ct +

+ 7)

'

' • « + ~ \

( < : < - ~ ) '

=

su\

1

oos

1

_- ?;

( , · ~ - T ~ + l , 1 _ , _ ~ - ~

- ' '' 2 <m - -

~ _ ( + ~ ' ) [ ( -')- l a + ~ + Z 7 ' I ]

~ S L n

2

co;

2

co\

2

;

2>in

( ;

2>in

( ;

r ;

i n : : ~ ;

Y)]

- 1"-')'

( I , ("")

" ' " ' . , 2 ' \ 2

:ill '

2

S i n c e , a , ~ , r c ( o .

=> ~ + {

"/+<>

E(O - '

2 2 2 - · ~ ~ )

= ;

sin I

< __:1-__ll:I

sin r ) _ ~ _yl sin p:_:i- -"]

'

0

,2) ,2 2

=>4,in

loin

s n : >0

HO'

I"'']

lpl

.2; 2 2

=> ' ' + . < i n ~ +>ill y- >ill (o + ~ + Tl> 0

,;,

a - siJ1

ll

+sin y> >iH (u

+

+ y)

6. AI X " 0,

l o g ~

,

"1U

X

and

Jog,"'

x

00{ X IS

llOI

Jcf.ncJ_ Sn, we

do

not determined ;he j {x)a(

x=O

9. TI1e equation or g:v"n """ '""

x - y ~ , ; < -

On differentiacing

w_u_

x,

we

get

" (

'

+

d . < ~ < '

y .xd .<

___ (i)

y

=

mx +

-; ,.

l

It

touche>,

x ' =

4ay. then

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n.

. ' [' "'""" v = 4<• mx +

-- 1

ha> equal roots

' '

:. -4<1m

2

x - 4o

2

=

0

I'"'

e<ludl wut>.

=

16a"m' +

l6 a

0

m = 0 (·: discrimi.,ant = 0)

m

1

=-1

m=-1

On puct;ng

m

=-1

my = m x-

-; ,,we

geL

y = x ~

= x + y ~ a = O

The "'quircd

p•ob.>bili y

= I - probab ley of each receiving ar least one

-1-n(l : :)

1t(S)

- ~ o w ,

the number of integral <olu.,on< or

x

1

  x

2

  x , - x

4

 X iX

 

olD

sud1 thatk

1

x, ?1, . . . .

,

g1v., "(E) ond the r.umh<.T

ul

mtegral>olutioos

of

x, .<,·>-

+-<,--' ,='n h iha'

x

1

20_x

2

l0 , x , _ ~ O g ' - ' ( S }

The required

pmbabll;ty

' ' c

l

,_,

=I

'"'''c,_,

6 137

- 1 4 1 = 1 4 . ~

'c

__l_

"c,

12.

for

ony x l, 2 we find t lm

f (x)

is d>e

quouent of two polynomials and a

f'Olynomial<

;, everywhere

continuou<.

Therefore, J'(>)"

rontinuou.< for o i l ~ - ~ I 2,

C011tlnull:y a t x

1,

LHL= lim f(x)=l i rn j( l -h)

, _ , - ' a

=

l•m (1 -h -2 ) (1 -h ;

2_1 1

-h + l)(J -h_-1)

.1-o

10-h

1)(1

h

2JI

=lirnl3-l t l(2-hlhU•+U

6

IHO h(h+ ) -

So _f(x)i' not conTinuous

at

x l.

Similor)y, _ftxl 1s

not conllnuous

at x

= 2.

Heru.,,

it

is

mn6nuau' for

R

- {l, 21-

1

~ ' '

mce, - - - - =o o ~ l + x )

(X·

2}(x

1

t-Il

+/,.,,-,_,-,

l l o ~ l x 2 l c

0

On

dil'fcrcrrtiating

both

sid.,, we get

1

d r ,

"· ' [ 'a lqdl +

x·)

(x+2)(x'+l) "-'

btan-

1

x+ loglx 1 2 >

tJ

2axb

---·---·---

+x '

l+ x

1

5 (X·I-2)

10ax(x+ 2)+ Sb(x + 2)+1 +

x

5(x+2lU+>- ' l

(x+2)(1+x'J

(l+lO.>)x

2

l ( % ~ 2 0 a ) : < + 1 i 1 0 b

' i ( x -2 ) (1+x

2

)

=<

(I

< IOol x

2

~ ( 5 0 + 20u)

,,

+I+ lOt=:;

On compar;ng the roe(flc.ent

of

x .

x

aTid

<;OTI> >:Tit

ttmO> on

both

sides, we

get

1 +lOa .'ilo-

20<1

o, lOb-4

j

2

""

a ~ - 1 0 , b ~ 5

14. Gi,..,n

1har, dy

=

r_

-

cus'

( ' ]

d:< X

X ,

Pu'

~ = 1 - - > _ ' f ~ x l ~ : t ~ r + x ~ ~

l +x - t = l - co s ' t

'ec·"t

dt -- -

'

.. im.-grnti[l)l,

we get

t :mt=-l<>g(x)+c

----;

tm :·I J =-log

(x)

+

c

'

(

Al p

uint

I,

:

I

Page 18: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 18/21

l

,,

Lan

"4)=

-log

(ll+c

 

c= l

= log

, )

 

an

l;

~ l o g

: ; : _

• l o g

tan(fJ=lo<;J

y=

xto

n· llo

g(fJ

]

15

L

enhe o

quoM

n of

 an y line

 throu

gh th

e on

gin

(O

,Q )b

oy-m

x"O .

 

.

..

(iJ

 

It

 

is

gi; c

n t ha

t

th

< le n

gth

of t11e

fr

om   rn,

 Pl LO

  li uc

(i) i l d.

l_Jt-

  :-J 

1

6. N\l

W, l«H

 

S tate

m ent

II   l<

th< w

rroa

axpla

n•tion

  o

f

Statement

L

17

Slo

peo

fim•2

xfy+

6=0

i.-2"

m  

(s

av) 

an

d slope

 of lin

e x -

2,v

-

5

 = 0

m

,  (sa

y)

m,m

, - - 1

 

A

lso.

Ll, ) l i e

on

x-2y

  5 =

0

H•

.nce, o

p t ion

 (d) is

 corre

ct .

18.

Since

, P"''

iV<I  of."

'· ' is

 21t.

'" 

~ 2 0 0

·. S

Mcm

ontl"

' false

  and

T

i ; 

true

.

3 + ~

- +

9

)

·

x

 x

 

hm

---

• -

4

+_2

__2

x'

x· 

3  0

-0 

0 3

0 -0

4

_

S Otom m t I

is

n11e

S ta t eme n t

IT

is 

true 

bur

ir  is

a  m

rrec r

  - P

lanati

on   of

 

Stat

emem

 T.

We

know

 tha

t equan

on o f

th e ll)

'll('tiv>

l' d ffe

i> 

l

om  ti l

e oqu

•t ion

uf  th e

  O'J'-

rnptme

  onl y

  in

oo n< t

on l te

rm <.

E

qua ll

on of

 

"

'")mp

to<e is

(x + 2

y + 3

)(3x-

4y

- 'i j =

0

_

Fqu

arion 

ofhyp

crbol

a

i ;

x + 2 y

< - 3 )

  3 x <

- 4 ~ •

+ 5 l +

A - O

, i)

lt

P"''"

' tl   ro u

gh   th e

 poim

 {I, -

1).

l

+

2   -1

)-

4(-

1 ) + 5

] +

  Q

'r om Eq.

(D

,

l>::o 

2y_,_ 3)(3_,

_, - -y

 

+

  5

- 8 ~ u

 

'43

x ' +

e J 

,

 

O >y

+ l 4x -

22y

  ·>7

~ o  

2

1.

L

et the

 c''"

" numb

er of

tefil'-

'l in G

P be 

2n, w1ti1

 

' ' ' t

( fm  

o ond

  oomrr

tGn ra

tio r

Then,

 

~ u r o

il rer

m ') (S

w:n

of

odd te

rms) 

=a,_ •

  0

2

 

+ •

•• -c«,

0

  .'i(<l.

, I

  J +

...  + <

t, ,_Jj

= a

a:r

2

_

.

+a

r' -'

"

{• '" -1

-

  r-1)

 

ar +

 ...

5-:I(r'"

 -

I

c

r I

5

r=

4

x-

1   y

2

"

E

ouau

on  r

lme>

ore

and

y   1

Z 2 

2 ----=

~ - 2

.

f

.quo l

on  ol

pl ane

pa,i

ng

h m t ~ g {2, -

1 -3

) is

a x -

2 ) + h

• : y +

l ) + c

  • + .

1 ) ~ 0

j

-<ow

giwn

 

l

ine5 a

re

pa

m id

 to

"-

3

a+2h

-4<-

-0

... [ll)

a

nd

J

a - 3b +

2<

• 0

... (Iii)

 

El

imina

tion of

"· ba

:nd cf

iorn 

E

qs_

i) , >i)  an

d

(iii),

we get

' ~ ~ q u o l m . r

,_

'0 -

Page 19: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 19/21

Now,

' ' 'Co

' +

1

C

2

  _

1

, '· C

1

=

'- c,., ...

'

1

C, =

'

c ,_,

·1

'c +' ·'c +. + '

1

C z'

~ - - '

--·---- =

(nt-1)

n+ l

21.

Gillen, 1

+II

-11

(c' -2)

=0 2+1< '-ll=, -2o'ol

- -- 2+le' - I H e

- t l

le -II'

-le'

-11- 2=0

l ie '

- l -2HI

e

-11

+ 11= o

le - l i=2.- l

e - 1 1 ~ 2

( ' le' - l loc-1)

d -1=2,-2

e' =3;-1

< =3

(·:<f;<-1)

_ The Tiumber ofreai ;ulull<ln ,;

l.

lx 3 7

Givel:tthat,

b

2=0

6 -

A p p l ) ' i n ~ R,--->

R

1

<

R,

1

R

3

and taking

rommon from

i

' '

(Xt-9)12

'

2 =0

:7

'

•I

Appl) ingc,

.... c , - c , a ~ d c , - - - > c , c,

c

,_,

_,

'=O

x-7 :

J x+ 4y= S

2;3x'i ,

3(4y l=5

,_2)

, 3 ,

3x 3>:

4y 4y

4y

2 2 3 3

7

3=

5

U•mg we1gl1Ced

AM

G'<l inequality

3.< 4v

1

2

~

3

~ [ ( > ; ) [ 4 ; ) )

L<>lZ=X+IJ

2ot- l 2 ( x + ( y ) + l _ • < ' ~ ' ; : ~ ' j l + i ( 2 y )

.. t z + l

=i ( x +[ v l+ l - - ( 1 y )+ i ( i )

~ ( 2 < 1 1)(1 v)+

2..w

]

_. +i {2y( l

y)

:.:(:1.>:<1)}

--- -( y ) + x

- l 2z+ l )

Smce, Tm•-,- - - 2

\'- +

1

( -2x

2

Ly' Y+2y)

l 2 y + x · y

W

e

ha

;

C

 

'

a

 

b

 

c

 

Page 20: JEE Sample Paper 5

7/23/2019 JEE Sample Paper 5

http://slidepdf.com/reader/full/jee-sample-paper-5 20/21

S

ll

l

0

5

° 

si

3

0"

 

>i

n 45

°

 

,

 3

+

1

 

- -

,

-

-J

3

+

I

 

2

si

n 1

0

5

°'

 

,1

3 +

 

c

 

~

 

,

1

si

lO

S

A

re

a

 

B

C

 

"

 1

 

si

~

~

 

e si

10

0

 

l

 

[

,.'3

 

l

l '

 

2

  u

z

,

in

(

6

0

°

t4

5

 

J

 

(

,• '

3

I l

f

 

4

.'2

1

''

3

 

i

I

1

 1

 

2

~

2

-

 

2

2

 

(,13+ 

ll  

2

(

./

3

+

1)

 

l

)  

t

an

 

[

.

] x

 

-

ta

n

 

r-

 

lx

1

sf

ii

2 X

 

'

-

ta

n

7

x

 

+

ta

n

a

x

 

m

 

<

in

7

 +

 

8

[

·

:7

<

e

2

<

8

)

 

3

1

.

L

e

t t.':t

c

d

io

m

o

:e

of

 

t

he

 

s

p

he

re

_

 A

E

 

2

Let 

the 

r idius 

of 

en 

no

"

 

_, 

a

nd

 b

e

ig

h t

 is

y

.

 

A

D

=

 

Y-

Sm

c

e

,  B

D

 

-A

D

-

D

E

 

x

 =

y

(2

r

-

y

)

 

__

_

( )

 

VO

 

W

lle

 o

co

n

e ,

 V

 

j

 

'

y  

[

fr

o

m

E

.q

.(

i)

=  ~

 

(

2

J

y

 -

y

 )

 

On 

dif fe rentiating w.r.r. 

y, 

we gor 

F

o

r m

ax

im

a

 

an

d

 m

in

im

a

.

p

ut

 :

 

=

 0

 

.

.,  

~ ~

(

4

r

y -

3

r

'

J

~ o

 

~

 

y

[4

r

-3

y

)

=

O

 

A

g

aJ

 l

d

if

fe

re

n

na

tl

n

g w

.

r.t

. y

,

 

w

e g

e

e

o

'v

 

1

 

J

y

,

=

3

r r

(4

r

-O

y

)

 

=

-v

e

 

V

o

lu

m

e

 o

c

on

e

 

is

 

m

ax

im

um

 a

t

 y

 

r

-

N

o

w

R

a

 

,

H

ei

gl

1t

 

o

f

 o

on

e

 

n

o =

D

ia

m

e

te

r

o

f -

<

ph

er

"

 

"

3

 

<---cor·

"F'I"t-1 

=4 i-J -3

k

 

Page 21: JEE Sample Paper 5

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> =

></

X •

tt /

2

I

 

i

n-

<d

x

ro

s

xd

x

 

14

x

]

f

' - I

sm

 

xJ

:;.

1

)

 

,

,(

 

-

33

S

in

ce

,

y  

=

  p

  x

)

O

n

 d

ll

fer

en

ti

ati

ng

 w

.r .

t.

x

  w

o g

e

t

2

1

 d

l=

p

'(

x

)

~

 

n

 og

a

in  

d i

f fe

, .

nn

a t

in

g ,

'

g

et

 

+

 2

y

=

 

p

"(

x)

 

2y

 

=

p

  {

x

)-

2

($

)

'

, .

  2

y

'

~ ~

=

y '

p

'

( r

)

- 2

y

'

( ~

)

'  

O

n

  ag

a

in

di

ffe

re

nt

ia t

in

g,

w

e g

et

 

(r 

1

 

= [

l(

x)

 

[f

'( x

)

+ p

(x

l

(  

' '

(x

x

dx  )  

- [

f(x

)

(.

<

=

P

lx

l-

f/

(

x)

 

,

l

li

xb

i

=-

J

i6

+

1+

Y

 

-

.

./2

6

.-. 

ni

t

Y<

ct

or

 po

rp

en

di

cu

ld

r t

oi

a

nd

 

il .

xb

 

_

_

 

1.1

 xD

I

 

i  

l-r41 

.,J-3i<J 

"

'

 

.

.

 

B

u

t-

-

,-

- (

4i

-

 

:lk

)m

a

ke

au

 a

cu

te

 a

ng

le  

,•2

6

 

w

ith

 the

  v

e<

:ro

r k

 

 

• 

Ro

qu

rr

ed

 v

ec

to r

 is

 --

-

 

- (

 

4

i -

- 3

k

).  

,

,

 

35

.

T

he

re

 11

re

30

 d

af

'

in

Ap

ri

l.

n(S

) =

th

e

n

u

mb

e

r of

 w

ay

in w

h

ic

0

 p

e

rw

ns

 

can h

av

e

in

 th e

 m

o

nt

h o

f A

p

nl

=

 

3

U

3U

 

x ___

_

10

 

t

im

e

s=

 30

·

0

(

be

ca

us

ea

ch

 p

ers

on

 c

an

 h

av

e b

in

 >

<la

in

 an

y

 

o

f 3

(]

wa

ys

).

 

n

(E

=

 

n

 S

t

he

 nu

m

be

r o

w

ay

in

 w

hk

h

 

10

 pe

r

n

'

C>

ln h

aw

. dif

fe

re

nt 

b

ir t

hd

;, y

s  

=

3o

'"

-

""

c"

 

» :. 30 

C

3