JEE Questions Definite Integration

7/24/2019 JEE Questions Definite Integration

1/17

J

l

1

f

i

L

L

I

N

T

H

E

B

L

A

N

K

S

'

;

t

t -

'

C

O

S

X

l .

c

o

;

' x

s

e

c

-

x

+

c

o

t

x

co

;

.e

c

x

I

o

o

s

e

"

'x

.

..

'

I

C

O

'

.

_

w

s

x

1

2

.

T

h

e

in

t

e

g

r

al

J

'

[

'

]

d

II

m

i

be

c

-om

in

u"

u.

' t

i.n

ct

ion

,.

Ti1cn

the v alue

of

he

in tegral

J

f x

)

+

f

x

)]

[g

(x

) -

g(

x

]

er

1>

l

hg

(2

)

:

os:

(

a)

31

2

{b

) 5

'

c

)

3


3/17

'

m

u

lt

ip

ly

(I

) b

y

a

a

2

f

(x

)

+

a

b

f(

l

lx

)

=

a

C

-s

)

m

ul

tip

ly

(

2

)b

y

b

a

b

f(

lt

x)

+

b

(

x

)

= b

(

x

-

5

)

S

u

bt

ra

c

t (

4)

f

ro

m

(

3)

(a

-

b )

f(x)

=

._

-

bx

-

5 1+

Sb

'

~

/

x

)

-

l

,

[

_-

b

x

-5

a

+

5

b

l

(

a

-

b

)

X

?\

low

,

f

'

'

'

[

"

l

x

d

x

=

-

-

~

-

b

x

-

5

a

+

5

b

d

r

I

(

a

-

b

)

I

X

-

(

3)

.

..

(4

)

a

l

o

g

l;

r

l-

-

x

-

s

(a

-

b

)x

'

[

b

l'

(a

2

1

)

2

1

'

[a

l

o

g

2

-2

b

-

IO

a

-

b

)-

a

lo

g

l

a

1

b )

L

ei

=

t

0

si

n:t

o

x

+

c

o

s'

x

-.

(1

)

l=

f

'

(1

1

t-

x)

[s

in

(

2:

1

t- x

)

J' "

d

r

0

[s

in

(2

:r:

x

)]

""

+

[

co

s

(2

,.

x)

]z

,

[

: ,C

J

(x

)d

B + 2 A = ~

7 r 2 ~ t

B

=0

Therefore, (d)

is

the answer.

9.

It

;, a question of

greate>t

integer function. We have

subdivide the

intervaln to

2n

'under

keep;ng in view

that we have tu ~ - v a l u a t e [2 sin x 1

'

~

-\ ,3, '2

-1/?, thi : '

Hence we divide the interval :tlo 27t as

(

' '1(7 lh:'jltlln

l

t '6,

6'6] -6 ' '

'"'o(o

l}('

l)(o i)

2sinx

=

0, - \) , ( -2

-1),

(Q - 1)

[2sinx]=-l

l = f ,+ l ,+1 ,

=

J

- l d n

J

2d:en proper limits

- - ~

-2(

:" - ~ - - 1 ~ = -

-:

Therefore, (a) is

the answer.

l()_ Applying

u.

I

j x)dx=

j" f ~ + b - x ) d

t,

2

Hence (c) is the amwcr

g(x)= ~ c o s 1 d l

).,

. , 4

g(.t+:r)= 0

cos tdl

(givcu)

J

" ' " J ..

'

_

0

cn.< lu i

co

dt=dy

"

= J ~ ( - o o , y ) d y =

J ~ c o s

yd>-=x(.d

g x

+

rt) =g{xH

g 1


6/17

j ( x )

.

l

+

I 2

T

he re

tim :

. (a)

is

tl

te an"

""'

,.,...

dx

j

n''

]+C

OS.

t

j

'

'"

"--

-c

=

14

x

)

. .( 1

)

(.

)

0

f(

x)d

x=

)

1

j

(a+

b-

x)d

x)

"

"

A

.dd

tng ( 1)

an d

(2

)

21=

+

J

y

(

' ' l

l'

]+co

sx

lCO

SX

v

~

J

'

I

-'-

J'

14

j

3rr.'4

l

z r - ~

m

11

sin

2

x

I

'

'

2

,

1=

w

se

,,.,

[

'

]

-cm

4

+

co t

4

l= -

(-1

)+1

=2

T

here

f


7/17

CJ(

x)d>

:=

C

,J(x)

dr+

f:J(

x)dx

J

'=

_

L

I '

_e

smxu

.t +

2

2d x

~ 0 + 2 [

. ~ J l

( :

sinx

is

a

n odd

fu n

ction

)

ff

(x)dl:=l

The

refore

. (c) is

the

answ e

r.

1

, J c

os'

x

,

1x;

=

oos

2

--

g r

r

duces

to

' "'

"'

6

1

f(y)

dy=

1= 3

1

1

t2 {

(

l+x

'i'l

L\

''J '

'" ,

_,,

,"'

J

'"

[I

+ X

I

[x

}dx+

log

-J '

-1 .

'2

1 -

x

'

J

(x ]d

nO

-1 1

2

.

In

S an

o

dd fun

c to n

>

I

(

(H

),

,

J

1-

x

J

,

j ,

[x)d>

: j''

2

[r ]d

O

whic

h oh

owsb

< 0on


8/17

i=( tdl=-xdx)

I=

(sin-

-sin

0)

+[tJ?

=.': '-].

'

7.

Jere,

I'

2

S

I

xj x)dx=-1

'

(using

Ne,. on

Leitonitz's formula};

diffenm iating

both

sid, we get

r '

{ / r

2

}} (r '

>)

- O . j ( O ) ~ .' _ 0})

= 21'

1dt ,th

t

2

f ( t

2

)2t-2r '

j ( r )= t

~ ) - - ~

' ' ) ' ' .

\zs

=s

(neg ectmg negat1ve)

' ~

I

= J ~ [ x

+ 3x' +3r+) +

x + }cos

(.x + l)jd>:

=

Ct:

D

SU JECTIVE QUESTIONS

,_

lim

[-'--'-+

+_ _]

.... n+l 11+2 2n

'

when

when

Similarly

.

""

ll _,-'

-'--"]

..... , ,

+ . 1

"

' '

n

' '

l m ~

, ...

, , = , r'l

-

, H )

'

'

=x

and

-"'dx.

'

=l , .x=-40.

=n.J


9/17

I

z

r

n

1-

I o

o+

UI S

/

1_

l l

) '

v

x

p

--

-

)

-

,

'1

,

CJO l

c.f I

.[

: S O ~ X

rj

(

'

j o

'

~

]

i S o o

lJr

- J

nsoo

x

J

,_)

p :ut

U S

' - +

.Y p

( ;ru

lllS

.\"1

/

Co

I

'

>>""

>]

'

:t

>[-

'YJ." UTS

Yl

'ru

.

ntl

li'X'

.

'

rp

)mu

;:xl

, . j

= r q

)

'"

')

'

')

cp

(xu S

)/ .

- rp

(n ns

) jx

"

"

. u

x p ( x

m s ) J

~ : J =Ii

:

; : ) p

(T )

.

)

(xp

(x-n

JfJ

~ x

p ( x )

f , jm

sn)

~ [

(I)

..

(zJ.

xp(

xu")

j(x-

lt)

")

~ ;

P(Y

111

>)fx:

J

J ' l g

( ~ )

\J (

]H l )

,.

xp

x

1

__iX

- [)

1

j

1+u

")

~

1 ~

"

(x-l

) }

"

l

. ,. ' '

.

1

""'"

c

'"

ll ,_

,( r- I

) ' .J"

+

(X

i) . .

Jx-

iJ'.J

.+ .(x-j

)O:)

J'

j

'

X ->lnXCOSX

9. Jl >1

fPCt- l ) ( t -2)

1

+3(r- l ) ' ( t -2J ' Idt


10/17

' ' J .

I

r

2 ~ _ 2 ~ - - ~ ~ m

0

cos''

x

+sm

' x

' j ' ' ' SlllXOOSX

1=-

2 sin

4

r+cos 'x

-

r/2 .ninxwox

d

Jo 1

csctsinx

Adding (I) and (2), get

21 = j

-- ' ;-;;;;c

o

1-cooasinx

1-tan' d:t:

'

l=1Cj

D l- I

(l+tan

-)-2cm,atan

' '

2 I = r

Zdt .wherct=hrn::

l+t +2/rosa 2

21=2

r

d

,

0

( r+cosar

+sina

=

sin a

[tan

'(' ~ : a

JI

=_;_ tan

1

(>)-tan

1

(cota))

sma

I '(.


11/17

I

l=

' , f ( t ) /(2a .

2.t +COS 4.t

+COS6T

... coo

(2k

- 2)

x}

co,2h

'-+2--

alu c

ofx

a

ndF

(x) ;,

co n t

in u ou

s .

:. F

(x) -O

for so

me

val

ue of x E(O,

b) o

r equ a

tio n ( l)

i

solvab le for.\.g

ain

f( -2

)

=

Q

(g iven

)- There

fore.

r

s 4

'J

t-

2)=

a

- + -

+ - H

l

3 )

3

(

-8

+1-

2 \

=>

O

=a

3

' )

'

-2

a

2

a

=>

c=

'

'

I

I

a

J

,

_

, _ 14

=>

- x

x

-

~ J =

= -

-

: 3

3

I

I

a

'

-(O

+x-

+

0+2)

d


13/17

-L

-J

=( [l e' l

-n

rr

dt J

Integrating b

y

p arts

by taking

1 l iS first function

= (

0- iIJ -

1

e-

1

-

nr(-

1

e dr

~ ( - 1 ) ~

-ne

j _r -''

dt

I . =(- J"-

1

-n/ ,_

1

lOr

n= l,

/

1

= f

> (.t - l )dx = [e

'(x - I ] ~ -

J> dx

= e

1

(

- I ) - e (O

- I )

- [e ' l

= l - (e

-1)

=2-e

Th er

efore, from equa

tion (1 ), we get

r ,=(- IJ ' ' -21,

=-l -2(1

-eJ=2 -5

and

1:, =(-1)

3

+

1

-.1/

2

=l

-3(2e-5)=16

-6e

Hence, n

=

is th

e answer.

19.

l=

j

2x

'+x

4

-2x

3

+2x

2

+1dx

2

( r '

+l)(x

4

)

J2x'

-2x

1

+ r

4

+1+2x

2

J

'

( x '-

I ) (x ' - l ) ( r

2

-

l )

= j'2x

1

(x ' - ) + ( ~ +1)

1

2

(x

2

-1)

2

(x

2

1)

'. (1)

- j ' 2x

3

(x

2

-l)

-- JJ

(x

2

+i)

2

- --

-=

rl

0

(x

2


14/17

l

+

1

:

sin xld

x

j

+ ,.

:t

2+

l> i

nx ldx

L

= 211

+ 1 - C

OS\'

,,,,

I

,.

I

,

J

Let

{= _

, , , \

1

-x

4

'co

s \ l

~ x

d

x

. 1

1

J

Pm

x

=

- _y so

thal

dx =-

dJ an

d

1

1 ,3

y

[ _,,

1

,

m

;

(-1

),zy

'

'

4

'

-

..

.,

J y

l+

y'

Now_

cos_

, (-x ="

- w s

- l

x lb

r-

I

,;

x

,; l rh

crefor

c,

;

'

i [

_,(

, ,

Jl

l t-C

O

=

> I

=

J

, e

w>

Documents

JEE Questions Definite Integration