QUESTION BANK ON
DEFINITE & INDEFINITE
INTEGRATION
XII (ALL)
M
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Time Limit : 5 Sitting Each of 100 Minutes duration approx.

Question bank on Definite & Indefinite Integration
There are 168 questions in this question bank.
Select the correct alternative : (Only one is correct)
Q.11/def The value ofthe definiteintegral, 





1
1
x
3
1
x
dx
)
e
e
( is
(A*) 2
e
4

(B)
e
4

(C) 






 
e
1
tan
2
e
1 1
2 (D) 2
e
2

[Hint: I = 



1
x
3
x
)
e
·
e
e
·
e
(
dx
= 


1
2
x
2
x
)
e
e
(
e
dx
e
(multiplyNr and Dr byex)
put ex = t  ex dx = dt
I = 


e
2
2
e
t
dt
e
1
=


e
1
2
e
t
tan
e
1
= 




 


4
2
e
1
2 = 2
e
4

Ans. ]
Q.22/def The value ofthedefinite integral, dx
e
x
2
·
e
cos
2
2
x
2
n
0
x






 






l
is
(A) 1 (B) 1 + (sin 1) (C*) 1 – (sin 1) (D) (sin 1) – 1
[Hint: put
2
x
e = t;
2
x
e · 2x dx = dt ; dt
t
cos
2
1


=   2
1
t
sin 
= 1 – (sin 1) ]
Q.35/def Value ofthe definite integral 






2
1
2
1
3
1
3
1
dx
)
)
x
3
x
4
(
cos
)
x
4
x
3
(
sin
(
(A) 0 (B*)
2

 (C)
2
7
(D)
2

[Hint: Note that in 






2
1
,
2
1
, sin–1(3x – 4x3) = 3 sin-1x and cos–1(4x3 – 3x) = 2 – 3 cos–1x
hence f (x) = 3 sin–1x – 2 + 3 cos–1x =
2


 I =
2
dx
2
2
/
1
2
/
1




 

]
[Alternate: f (x) = sin–1 (3x – 4x3) – [ – cos–1(3x – 4x3 ) ]
= – + (sin–1 (3x – 4x3) + cos–1 (3x – 4x3)) =
2

 ]
Q.46/def Let f (x) =  
x
2
4
t
1
dt
and g be the inverse off. Then the value ofg'(0) is
(A) 1 (B) 17 (C*) 17 (D) noneofthese

[Sol. f ' (x) =
4
x
1
1

=
dx
dy
now g ' (x) =
dy
dx
= 4
x
1
when y = 0 i.e.  
x
2
4
t
1
dt
= 0 then x = 2 (think !)
hence g '(0) = 16
1 = 17 ]
Q.56/inde 

x
x
1
e
)
e
(
cot
dx is equal to :
(A)
2
1
ln (e2x + 1)  x
x
1
e
)
e
(
cot
+ x + c (B)
2
1
ln (e2x + 1) + x
x
1
e
)
e
(
cot
+ x + c
(C*)
2
1
ln (e2x + 1)  x
x
1
e
)
e
(
cot
 x + c (D)
2
1
ln (e2x + 1) + x
x
1
e
)
e
(
cot
 x + c
Q.69/def  

k
0
x
1
0
k
dx
)
x
2
sin
1
(
k
1
Lim
(A) 2 (B) 1 (C*) e2 (D)nonexistent
[Hint: l =
k
dx
)
x
2
sin
1
(
Lim
k
0
x
1
0
k
 

differentiating
Using L'opital rule l = k
1
0
k
)
k
2
sin
1
(
Lim 

=
)
k
2
(sin
k
1
Lim
0
k
e 
= e2 ]
Q.711/def  

5
n
0
x
x
x
3
e
1
e
e
l
dx =
(A*) 4   (B) 6   (C) 5   (D) None
[Hint: put ex  1 = t2 ]
Q.816/def If x satisfies the equation 2
1
0
2
x
1
cos
t
2
t
dt











 – x
dt
1
t
t
2
sin
t
3
3
2
2











– 2 = 0 (0 <  < ), then the
value xis
(A) ±


sin
2
(B) ±


sin
2
(C) ±


sin
(D*) ±


sin
2

[Sol. 
 
3
3
2
2
dt
1
t
t
2
sin
t
= 0 as the integrand is an odd function.
also  


1
0
2
1
cos
t
2
t
dt
=
1
0
1
sin
cos
t
tan
sin
1





=


sin
2
Thus the givenequationreduces to
x2


sin
2
– 2 = 0  x = ±


sin
2 Ans.]
Q.917/def If f (x) = eg(x) and g(x) =
2
x

t dt
t
1 4

then f (2) has the value equalto :
(A*) 2/17 (B) 0 (C) 1 (D) cannot be determined
[Hint: f (x) = eg (x). g (x) ; g (x) =
x
x
1 4

hence f (2) = eg (2). g  (2) = e0.
2
17
=
2
17
]
Q.107/inde  etan  (sec – sin)d equals :
(A)  etan  sin  + c (B) etan  sin  + c (C) etan  sec  + c (D*) etan  cos  + c
[Hint: integrate  etan  . sin  d byparts. One integralcancels.
Q.1118/def 

0
(x · sin2x · cosx) dx =
(A) 0 (B) 2/9 (C)  2/9 (D*) 4/9
[Hint: 

0
2
dx
)
x
cos
x
(sin
·
x = 



0
3
0
3
dx
x
sin
3
1
3
x
sin
·
x ]
Q.1222/def The value of
 






n
4
r
1
r
2
n
n
4
r
3
r
n
Lim is equalto
(A)
35
1
(B)
14
1
(C*)
10
1
(D)
5
1
[Sol. Tr = 2
4
n
r
3
n
·
n
r
1









S = 









n
4
1
2
n
r
·
4
n
r
3
1
n
1
=  
4
0
2
)
4
x
3
(
x
dx

put 4
x
3  = t  dt
dx
x
1
2
3

= 
10
4
2
t
dt
3
2
= 













10
1
4
1
3
2
t
1
3
2
4
10
=
10
1
40
6
·
3
2
 ]
Q.1323/def 



c
b
c
a
dx
)
c
x
(
f =
(A*) 
b
a
dx
)
x
(
f (B)  
b
a
dx
)
c
x
(
f (C) 


c
2
b
c
2
a
dx
)
x
(
f (D)  
b
a
dx
)
c
2
x
(
f
[Hint: Put x + c = t ]
Q.1425/def Let I1 = 



2
/
0
dx
x
cos
.
x
sin
1
x
cos
x
sin
; I2 = 

2
0
6
dx
)
x
cos
( ; I3 = 



2
/
2
/
3
dx
)
x
(sin & I4 =  






1
0
dx
1
x
1
n
l then
(A) I1 = I2 = I3 = I4 = 0 (B) I1 = I2 = I3 = 0 but I4  0
(C*) I1 = I3 = I4 = 0 but I2  0 (D) I1 = I2 = I4 = 0 but I3  0
Q.158/inde  

)
x
1
(
x
x
1
7
7
dx equals :
(A) lnx +
2
7
ln (1 + x7) + c (B) lnx 
2
7
ln (1  x7) + c
(C*) lnx 
2
7
ln (1 + x7) + c (D) lnx +
2
7
ln (1  x7) + c
[Hint: I =
 
dx
x x
1 7

 
x
x
6
7
1 
 dx ]
Q.1626/def 


n
2
/
0
n
nx
tan
1
dx
=
(A) 0 (B*) n
4

(C)
4
n
(D) n
2

[Hint: nx = t; I = 


2
0
n
)
t
(tan
1
dt
n
1
= 


2
0
n
n
n
dt
)
t
(cos
)
t
(sin
)
t
(cos
n
1
]
Q.1728/def f(x) =  

x
0
dt
)
2
t
(
)
1
t
(
t takes onits minimumvalue when:
(A) x= 0 , 1 (B) x = 1 , 2 (C*) x = 0 , 2 (D) x =
3 3
3


[Hint: f ' (x) = x (x – 1) (x – 2) = 0
x = 0, 1, or 2
at x = 0 & 2, f ' (x) changes sign from – ve to + ve  minimum ]
Q.1829/def 

a
a
dx
)
x
(
f =
(A*)  
 

a
0
dx
)
x
(
f
)
x
(
f (B)  
 

a
0
dx
)
x
(
f
)
x
(
f (C) 2 
a
0
dx
)
x
(
f (D) Zero
[Hint: I = 

a
a
dx
)
x
(
f = 


a
a
dx
)
x
(
f (using K)
 2I =  




a
a
dx
)
x
(
f
)
x
(
f =  
 

a
0
dx
)
x
(
f
)
x
(
f
2 ( as integralis even)
 result ]
Q.1930/def Let f (x) be a function satisfying f ' (x) = f (x) with f (0) = 1 and g be the function satisfying
f (x) + g (x) = x2. The value ofthe integral 
1
0
dx
)
x
(
g
)
x
(
f is
(A) e –
2
1
e2 –
2
5
(B) e – e2 – 3 (C)
2
1
(e – 3) (D*) e –
2
1
e2 –
2
3
[Sol. f ' (x) = f (x)  f (x) = C ex and since f (0) = 1
 1 = f (0) = C  f (x) = ex and hence g (x) = x2 – ex
Thus 
1
0
dx
)
x
(
g
)
x
(
f =  
1
0
x
2
x
2
dx
)
e
e
x
(
=
1
0
x
2
1
0
x
1
0
x
2
2
e
dx
xe
2
e
x 
  = (e – 0) – 2 [
1
0
x
xe –
1
0
x
e ] –
2
1
(e2 – 1)
= (e – 0) – 2 [(e – 0) – (e – 1)] –
2
1
(e2 – 1)
= e –
2
1
e2 –
2
3
]
Q.2011/inde   |
x
|
n
1
x
|
x
|
n
l
l
dx equals :
(A*) |
x
|
n
1
3
2
l
 (lnx 2) + c (B) |
x
|
n
1
3
2
l
 (lnx+ 2) + c
(C) |
x
|
n
1
3
1
l
 (lnx 2) + c (D) |
x
|
n
1
2 l
 (3 lnx 2) + c
[Hint: Start : ln | x | = t  dx
x
1
= dt    t
1
dt
t
]

Q.2131/def  
 









2
1
3
2
1
dx
4
|
x
1
|
|
3
x
|
2
1
equals:
(A)
2
3
 (B)
8
9
(C*)
4
1
(D)
2
3
Where {*} denotesthe fractionalpart function.
[Hint:  
 









2
1
3
2
1
dx
4
|
x
1
|
|
3
x
|
2
1
 
 












1
2
1
dx
4
x
1
x
3
2
1
+  
 











3
1
dx
4
1
x
x
3
2
1
+  
 









2
7
3
dx
4
1
x
3
x
2
1
= 
 


2
7
3
1
2
1
dx
}
4
x
{
dx
}
x
{ = 
 

2
7
3
1
2
1
dx
}
x
{
dx
)
x
1
( = 
 


2
7
3
1
2
1
dx
)
3
x
(
dx
)
x
1
( ]
Q.2232/def 

/
4
0







x
1
cos
.
x
x
1
sin
.
x
3 2
dx has the value :
(A)
8 2
3

(B)
24 2
3

(C*)
32 2
3

(D) None
[Hint: Consider 

4
0
2
dx
x
1
sin
x
3 and IBP taking
x
1
sin as the 1st and 3x2 as the 2nd function. Two integrals
cancel]
[Sol.

 









 dx
x
1
cos
x
x
1
sin
x
3
I
2
=
3
x
·
x
1
sin –  





 dx
x
x
1
x
1
cos 3
2 –  dx
x
1
cos
x 2
=

4
0
3
x
1
sin
·
x ]
Q.2338/def 















 






 


 3
4
n
6
)
1
n
(
sec
.....
n
6
·
2
sec
n
6
sec
n
6
Lim 2
2
2
n
has the value equalto
(A*)
3
3
(B) 3 (C) 2 (D)
3
2
[Sol. Tr =
n
6
r
sec
n
6
2 

S = 





n
1
r
2
r
n
6
r
sec
n
6
T = 


1
0
2
dx
6
x
sec
6
=
1
0
6
x
tan

=
3
1
=
3
3
]

Q.2439/def Suppose that F(x) is an antiderivative of f(x) =
x
x
sin
, x > 0 then 
3
1
dx
x
x
2
sin
canbe expressed as
(A*) F (6) – F (2) (B)
2
1
( F (6) – F (2) ) (C)
2
1
( F (3) – F (1) ) (D) 2( F (6) – F (2) )
[Hint: Given  dx
x
x
sin
Let 2x = t, dx =
2
dt
 I = 


6
2
dt
t
t
sin
2
2
= 
6
2
dt
t
t
sin
But  dx
x
x
sin
= F (x)  
6
2
dt
t
t
sin
= F (6) – F (2) ]
Q.2513/inde Primitive of 2
4
4
)
1
x
x
(
1
x
3



w.r.t. x is :
(A)
1
x
x
x
4


+ c (B*) 
1
x
x
x
4


+ c (C)
1
x
x
1
x
4



+ c (D) 
1
x
x
1
x
4



+ c
[Hint:
 
3 1
1
4
2 3 1
2
x
x x x

  
=
 
3
1
2 2
3 1
2
x x
x x

 


]
Q.2641/def 

n
Lim    
2
1
2
2
2
1
2
n n n
n
n
   







cos cos ..... cos
( )
equal to
(A*) 1 (B)
1
2
(C) 2 (D) none
[Hint: Tr =
n
2
r
cos
n
2


S = 



 1
n
0
r n
r
·
2
cos
n
1
·
2 = 


1
0
dx
2
x
cos
2
= 1 ]
Q.2742/def
 
log
log
x
x
n
2
2
2
2
2
4









 
dx =
(A*) 0 (B) 1 (C) 2 (D) 4
[Hint: dx
x
n
2
n
nx
2
n
4
2
2
 






l
l
l
l
if f(x) = x
n
1
l x
n
1
)
x
(
'
f
x 2
l



 I = ln2
4
2
nx
x






l
= 






2
n
2
4
n
4
2
n
l
l
l = 0 ]

Q.2844/def If m & n are integers such that (m  n) is an odd integer then the value of the definite integral


0
dx
nx
·sin
mx
cos =
(A) 0 (B*) 2
2
m
n
n
2

(C) 2
2
m
n
m
2

(D) none
[Hint: Note that ifm–nis odd m+nisalso odd.Split the integrandusing 2 sinA cosB as sumor difference.]
Q.2945/def Let y={x}[x] where {x}denotes the fractional part of x & [x] denotes greatest integer  x, then

3
0
dx
y =
(A) 5/6 (B) 2/3 (C) 1 (D*) 11/6
Q.3015/inde If
 



2
2
4
1
x
x
1
x
dx =Aln x+ 2
x
1
B

+ c , where c is the constant of integration then :
(A) A = 1 ; B =  1 (B) A =  1 ; B = 1 (C*) A = 1 ; B = 1 (D) A =  1 ; B =  1
[Hint: add and subtract x2 inthe numerator]
Q.3148/def 




2
/
x
cos
1
x
sin
1
dx =
(A*) 1  ln2 (B) ln2 (C) 1 + ln2 (D) none
[Hint: 

 dx
2
x
sin
2
x
sin
1
I
2
=  





 dx
2
x
cot
2
x
ec
cos
2
1 2
= –x cot

 2
2
x
Q.3249/def Let f : R  R be a differentiable function & f(1) = 4, then the value of;
1
x
Lim
  
)
x
(
f
4
1
x
dt
t
2 is :
(A) f(1) (B) 4f(1) (C) 2f (1) (D*) 8f(1)
Q.3350/def If 
)
x
(
f
0
2
dt
t = x cos x , then f ' (9)
(A*) is equalto –
9
1
(B) is equalto –
3
1
(C) is equal to
3
1
(D) isnonexistent
[Sol.
)
x
(
f
0
3
3
t
= x cos x  [f (x)]3 = 3x cos x ....(1)
[f (9)]3 = – 27  f (9) = – 3
also differentiating 
)
x
(
f
0
2
dt
t = x cos x

[f (x)]2 · f ' (x) = cos x – x  sin  x
 [f (9)]2 · f ' (9) = – 1
 f ' (9) = –
 2
)
9
(
f
1
= –
9
1
f ' (9) = –
9
1
 (A) ]
Q.3451/def 
 3
/
1
)
2
/
(
0
3
5
dx
x
·sin
x =
(A) 1 (B) 1/2 (C) 2 (D*) 1/3
Q.3516/inde Integral of )
x
ec
cos
x
(cot
x
cot
2
1 
 w.r.t. x is :
(A) 2 ln cos
2
x
+ c (B*) 2 ln sin
2
x
+ c
(C)
2
1
ln cos
2
x
+ c (D) ln sin x  ln(cosec x  cot x) + c
[Hint: I =  
 x
cot
2
x
cot
x
ec
cos
2
1 2
=  
 dx
x
cot
x
cot
ecx
cos
2
x
ec
cos 2
2
=   dx
)
x
cot
ecx
(cos ]
Q.3652/def If f (x) = x+ x1 + x 2, x  R then 
3
0
dx
)
x
(
f =
(A) 9/2 (B) 15/2 (C*) 19/2 (D) none
Q.3756/def Number ofvalues ofx satisfying the equation 









x
1
2
dt
4
t
3
28
t
8 =
 
1
x
log
1
x
)
1
x
(
2
3



, is
(A) 0 (B*) 1 (C) 2 (D) 3
[Hint: Integrating LHS a cubic in x ofthe formx(ax2 + bx + c) = 0
giving x=0, –3/2,–1/4. Onlyx = –1/4 satisfies]
Q.3858/def 

1
0
1
dx
x
x
tan
=
(A) 
 4
/
0
dx
x
x
sin
(B) 
 2
/
0
dx
x
sin
x
(C*) 
 2
/
0
dx
x
sin
x
2
1
(D) 
 4
/
0
dx
x
sin
x
2
1
[Sol. I = 

1
0
1
dx
x
x
tan
x = tan ; dx = sec2 d

I = 





4
/
0
2
d
tan
sec
.
= 





4
/
0
d
cos
sin = 




4
/
0
d
2
sin
2
2 = y
= 
 2
/
0
dy
y
sin
y
2
1
= 
 2
/
0
dx
x
sin
x
2
1
 (C) ]
Q.3961/def Domain ofdefinition ofthe function f(x) =  
x
0
2
2
t
x
dt
is
(A) R (B) R+ (C) R+  {0} (D*) R – {0}
Q.4017/inde If  e3x cos 4x dx = e3x (A sin4x + B cos 4x) + c then :
(A) 4A = 3B (B) 2A = 3B (C*) 3A = 4B (D) 4B + 3A = 1
Q.4163/def If f (a + b x) = f(x) , then  

b
a
dx
)
x
b
a
(
f
.
x =
(A) 0 (B)
2
1
(C*) 

b
a
dx
)
x
(
2
b
a
f (D) 

b
a
dx
)
x
(
2
b
a
f
Q.4265/def The set ofvalues of'a' which satisfythe equation  
2
0
2
dt
)
a
log
t
( = log2 





2
a
4
is
(A) a  R (B*) a  R+ (C) a < 2 (D) a > 2
[Hint:
2
0
2
2
t
·
a
log
2
t
 = 2 – log2(a2)
(2 – 2 log2a) = 2 – 2 log2a
2 log2a = 2 log2a  a  R+ ]
Q.4366/def The value ofthe definite integral dx
)
5
x
4
(
5
x
2
)
5
x
4
(
5
x
2
3
2
 




 



 =
(A)
7 3 3 5
3 2

(B) 4 2 (C) 4 3 +
4
3
(D*)
7 7 2 5
3 2

[Hint: Put 4x  5 = 5t2
 4dx = 10t dt or better will be 5(4x – 5) = t2 ]
I = dt
t
t
5
)
t
1
(
2
5
t
5
)
t
1
(
2
5
2
5 5
7
5
3
2
2
 



 =   dt
t
|
)
1
t
(
|
|
1
t
|
2
5 5
7
5
3
2
/
3
 








=    


























 dt
t
)
1
t
(
)
1
t
(
dt
t
|
)
t
1
(
|
)
t
1
(
2
5
5
7
1
1
5
3
2
/
3

=




















 dt
t
dt
t
2
2
5 2
5
7
1
1
5
3
2
/
3
]
Q.4467/def Number ofordered pair(s) of(a, b) satisfying simultaneouslythesystemofequation
0
dx
x
b
a
3

 and
3
2
dx
x
b
a
2

 is
(A) 0 (B*) 1 (C) 2 (D) 4
[Hint: a = – 1 and b = 1 ]
Q.4518/inde  





x
cot
x
tan
x
cot
x
tan
1
1
1
1
dx is equal to :
(A)

4
x tan1 x +

2
ln (1 + x2)  x + c (B)

4
x tan1 x 

2
ln (1 + x2) + x + c
(C)

4
x tan1 x +

2
ln (1 + x2) + x + c (D*)

4
x tan1 x 

2
ln (1 + x2)  x + c
Q.4668/def Variable x and y are related by equation x=  
y
0
2
t
1
dt
. The value of 2
2
dx
y
d
is equalto
(A) 2
y
1
y

(B*) y (C) 2
y
1
y
2

(D) 4y
[Hint: x = ln 




 
 2
y
1
y ]
Q.4769/def Let f (x) = 




h
x
x
2
0
h
t
1
t
dt
h
1
Lim , then )
x
(
·
x
Lim
x
f



is
(A) equalto 0 (B) equalto
2
1
(C) equalto 1 (D*) nonexistent
[Hint: f (x) = x
x
1 2

 ; 




 




x
x
1
x
Lim 2
x
 –   DNE ]
Q.4873/def If the primitive of f (x) =  sin x + 2x 4, has the value 3 for x = 1, thenthe set of x for which the
primitive off(x) vanishesis :
(A) {1, 2, 3} (B) (2, 3) (C*) {2} (D) {1, 2, 3, 4}

Q.4974/def If f & g are continuous functions in [0, a] satisfying f (x) = f(a  x) & g (x) + g (a  x) = 4 then

a
0
dx
)
x
(
g
.
)
x
(
f =
(A) 
a
0
2
1
f(x)dx (B*) 
a
0
2 f(x)dx (C) 
a
0
f(x)dx (D) 
a
0
4 f(x)dx
[Sol. I = dx
)
x
(
g
.
)
x
(
f
a
0
 ....(1)
I = dx
)
x
a
(
g
.
)
x
a
(
f
a
0



I = dx
)
)
x
(
g
4
(
.
)
x
(
f
a
0

 = 
 
a
0
a
0
)
x
(
g
)
x
(
f
)
x
(
f
4 ....(2)
(1) + (2) 2I = 
a
0
)
x
(
f
4  I = dx
)
x
(
f
2
a
0
  (B) ]
Q.5020/inde  x . 2
2
x
1
x
1
x
n






 

l
dx equals :
(A*) 2
x
1 ln 




 
 2
x
1
x  x + c (B)
2
x
. ln2 




 
 2
x
1
x  2
x
1
x

+ c
(C)
2
x
. ln2 




 
 2
x
1
x + 2
x
1
x

+ c (D) 2
x
1 ln 




 
 2
x
1
x + x + c
[Hint: use I.B.P. taking ln 




 
 2
x
1
x as the first functionand
x
x
1 2

as the second function ]
Q.5175/def If f (x) =





2
x
1
)
6
x
7
(
1
x
0
x
1
3
1







, then 
2
0
dx
)
x
(
f is equal to
(A)
6
31
(B)
21
32
(C)
42
1
(D*)
42
55
Q.5277/def The value ofthe definite integral  
1
0
x
e
dx
)
e
·
x
1
(
e
x
is equalto
(A*) ee (B) ee – e (C) ee – 1 (D) e
[Hint: start with ex = t and use  
  dx
)
x
(
'
f
)
x
(
f
ex
= ex f(x) ]

Q.5379/def  






2
2
/
1
dx
x
1
x
sin
x
1
has the value equalto
(A*) 0 (B)
4
3
(C)
4
5
(D) 2
[Sol. x =
t
1
 dx = 2
t
1
dt
I =  













2
/
1
2
2
dt
t
1
t
t
1
sin
t =  






2
/
1
2
dt
t
1
t
sin
t
1
= –  






2
2
/
1
dt
t
1
t
sin
t
1
= – I
 2I = 0  I = 0
Alternatively : put x = et  I = 



2
n
2
n
t
t
dt
)
e
e
sin(
l
l
= 0 (odd function) ]
Q.5481/def The value of the integral 

0
e2x (sin 2x + cos2x) dx =
(A) 1 (B) 2 (C*) 1/2 (D)zero
[Hint: Put  2x = t and use  ex ( f(x) + f (x) ) d x = ex f(x) + c ]
Q.5582/def The value ofdefinite integral 




0
z
2
z
dz
e
1
e
z
.
(A*) – 2
n
2
l

(B) 2
n
2
l

(C) –  ln 2 (D)  ln 2
[Hint: Put e–z = sin  ]
Q.5622/indeAdifferentiable function satisfies
3f 2(x) f '(x) = 2x. Given f (2) = 1 then the value of f (3) is
(A) 3
24 (B*) 3
6 (C) 6 (D) 2
[Hint: Integrating both sides we gets f3(x) = x2 + C ; f (2) = 1  C = – 3
f 3(x) = x2 – 3  f 3(3) = 6  f (3) = 3
6 ]
Q.5786/def For In = 
e
1
(lnx)ndx, n  N;which of the following holds good?
(A) In + (n + 1) In + 1 = e (B) In + 1 + n In = e
(C*) In + 1 + (n +1) In = e (D) In + 1 + (n – 1) In = e
[Hint: In = e – nIn – 1 ]

Q.5889/def Let fbe acontinuous functions satisfying f ' (lnx) =


1
x
x for
1
x
0
for
1



and f(0) = 0 thenf (x) can
be defined as
(A) f(x) = 


0
if x
e
1
0
if x
1
x



(B) f(x) = 


0
if x
1
e
0
if x
1
x



(C) f (x) = 


0
if x
e
0
if x
x
x


(D*) f(x) = 


0
if x
1
e
0
if x
x
x



[Sol. f ' (ln x) = 


1
x
x for
1
x
0
for
1



put ln x = t  x = et
for x > 1 ; f ' (t) = et for t > 0
integrating f (t) = et + C ; f (0) = e0 + C  C = – 1
 f (t) = et – 1 for t > 0 (corresponding to x > 1)
 f (x) = ex – 1 for x > 0 ....(1)
again for 0 < x  1
f ' (ln x) = 1
f ' (t) = 1 for t  0
f (t) = t + C
f (0) = 0 + C  C = 0  f (t) = t for t  0  f (x) = x for x  0]
Q.5991/def Let f: R  R be a differentiable functionsuch that f(2) = 2. Then the value ofLimit
x  2
4
2
3
2
t
x
f x


( )
dt is
(A) 6 f (2) (B) 12 f (2) (C*) 32 f (2) (D) none
Q.6093/def 


2
/
0
2
2
x
sin
a
1
dx
has the value :
(A*)

2 1 2
a
(B)

1 2
a
(C)
2
1 2

a
(D) none
[Hint: I = 



2
/
0
2
2
2
2
x
tan
a
x
tan
1
dx
x
sec
= 



2
/
0
2
2
2
1
x
tan
)
a
1
(
dx
x
sec
]
Q.6124/inde Let f(x) = 





x
e
x
n
x
1
l then its primitive w.r.t. x is
(A)
2
1
ex – ln x + C (B)
2
1
ln x – ex + C (C*)
2
1
ln2x – x + C (D)
x
2
ex
+ C
[Sol.  dx
e
x
n
x
1
x
l =   dx
)
ne
nx
(
x
1 x
l
l

= 

dx
x
x
nx
l
= 





 
 dx
x
x
1
dx
x
n
x
1
l (put ln x = u ; du
dx
x
1
 )
=  dx
u –  dx
1 =
2
1
ln2x – x + C ]
Q.6294/def 


 
n
1
k
2
2
2
n x
k
n
n
Lim , x > 0 is equal to
(A) x tan–1(x) (B) tan–1(x) (C*)
x
)
x
(
tan 1

(D) 2
1
x
)
x
(
tan
Q.6397/def Let f(x) =
2 2
2 2
0
2
2
cos sin( ) sin
sin sin cos
sin cos
x x x
x x x
x x


then
0
2
/
 [f(x) + f (x)] dx =
(A*)  (B) /2 (C) 2  (D)zero
[Hint: Use c2  c2 + 2 cos x c3  f(x) = 2  f (x) = 0 ]
Q.64100/def The absolute value of sinx
x
1 8
10
19

 is less than :
(A) 10 10 (B) 1011 (C*) 10 7 (D) 109
[Hint:
sinx
x
1 8
10
19

 
sinx
x
1 8
10
19

 dx 
dx
x
1 8
10
19

 <
dx
x8
10
19
 =
x







7
10
19
7
= 
1
7
[19 7  10 7] =
1
7
[10 7  19 7] < 10 7 ]
Q.65101/def The value ofthe integral




(cos px  sin qx)2 dx where p, q are integers, is equalto :
(A)  (B) 0 (C)  (D*) 2
[Hint : I = 



 dx
)
qx
sin
px
(cos 2
I = 



 dx
)
qx
sin
px
(cos 2
(UsingKing)
2I = 2 



 dx
)
qx
sin
px
(cos 2
2
I = 


0
2
2
dx
)
qx
sin
2
px
cos
2
( = 




0
dx
)
qx
2
cos
1
(
)
px
2
cos
1
(
= 2 (Boththe integrals vanish)

Q.6625/inde Primitive of f (x) = )
1
x
(
n 2
2
·
x 
l
w.r.t. x is
(A)
)
1
x
(
2
2
2
)
1
x
(
n 2


l
+ C (B)
1
2
n
2
)
1
x
( )
1
x
(
n
2 2

 
l
l
+ C
(C*)
)
1
2
n
(
2
)
1
x
( 1
2
n
2

 
l
l
+ C (D)
)
1
2
n
(
2
)
1
x
( 2
n
2


l
l
+ C
[Sol. I = 

dx
2
x )
1
x
(
n 2
l
let x2 + 1 = t ; x dx =
2
dt
Hence I =  dt
2
2
1 t
n
l
=  dt
t
2
1 2
n
l
=
1
2
n
t
·
2
1 1
2
n


l
l
+ C =
1
2
n
)
1
x
(
·
2
1 1
2
n
2

 
l
l
+ C  (C) ]
Q.67103/def  









2
0
n
n
dt
1
n
t
1
Lim is equal to
(A) 0 (B) e2 (C*) e2 – 1 (D) does not exist
[Sol.  









2
0
n
n
dt
1
n
t
1
Lim =
2
0
1
n
n 1
n
t
1
Lim



















= 1
1
n
2
1
Lim
1
n
n












= e2 – 1













 function
linear
a
is
1
n
t
1 ]
Q.68106/def
Limit
h  0
 
n t dt n t dt
h
a
x h
a
x
2 2

 

=
(A) 0 (B*) ln2 x (C)
2nx
x
(D) does not exist
Q.69108/def Let a, b, c be nonzero real numbers such that ;
0
1
 (1 + cos8x) (ax2 + bx + c) dx =
0
2
 (1 + cos8x) (ax2 + bx + c) dx , then the quadratic equation
ax2 + bx+ c = 0 has :
(A) no root in (0, 2) (B*) atleast one root in (0, 2)
(C) a double root in (0, 2) (D) none
[Hint:  
1
0
8
dx
)
x
(
f
)
x
cos
1
( =  
2
0
8
dx
)
x
(
f
)
x
cos
1
( = 
 


2
1
8
1
0
8
dx
)
x
(
f
)
x
cos
1
(
dx
)
x
(
f
)
x
cos
1
(
Hence  
2
1
8
dx
)
x
(
f
)
x
cos
1
( = 0
 (1+cos8x) f(x) = 0 at least once in (1,2)
but 1 + cos8x 0
 f(x) = ax2 + bx + c vanishes at least once in (1,2) ]

Q.70110/def Let In =
0
4
/
 tann x dx, then
1 1 1
2 4 3 5 4 6
I I I I I I
  
, , ,.... are in :
(A*)A.P. (B) G.P. (C) H.P. (D) none
[Hint: In =
0
4
/
 tann  2 x (sec2 x  1) dx =
1
1
n 
 In  2  In + In  2 =
1
1
n 
 I2 + I4 =
5
1
; I5 + I3 =
6
1
and I6 + I4 =
7
1
 result ]
Q.7127/inde Let g (x) be an antiderivative for f (x). Then ln  
 
2
)
x
(
g
1 is an antiderivative for
(A)
 2
)
x
(
1
)
x
(
)
x
(
2
f
g
f

(B*)
 2
)
x
(
1
)
x
(
)
x
(
2
g
g
f

(C)
 2
)
x
(
1
)
x
(
2
f
f

(D) none
[Sol. Given  dx
)
x
(
f = g (x)  g ' (x) = f (x)
now  
)
x
(
g
1
(
n
dx
d 2

l =
)
x
(
g
1
)
x
(
'
g
)
x
(
g
2
2

=
)
x
(
g
1
)
x
(
)
x
(
2
2

g
f
 (B)]
Q.72114/def
0
4
/
 (cos 2x)3/2. cos x dx =
(A)
3
16

(B)
3
32

(C*)
3
16 2

(D)
3 2
16

[Hint: I =
0
4
/
 (1  2 sin2 x)3/2 cos x dx. Put 2 sin x = sin 
 I =
1
2 0
2
/
 cos4  d =
3
16 2

]
Q.73116/def The value ofthe definite integral  


2
1
0
2
2
2
)
x
1
1
(
x
1
dx
x
is
(A)
4

(B)
2
1
4


(C*)
2
1
4


(D) none
[Sol. I =  


2
1
0
2
2
2
)
x
1
1
(
x
1
dx
x
put x = sin ;dx = cos d
= 







4
0
2
)
cos
1
(
cos
d
cos
sin
= 




4
0
d
)
cos
1
( = =   4
0
sin 


 = 







2
1
4
Ans. ]

Q.74117/def The value ofthe definite integral  
 

37
19
2
dx
)
x
2
(sin
3
}
x
{ where { x} denotes the fractionalpart
function.
(A) 0 (B*) 6 (C) 9 (D) cannot be determined
Q.75122/def The value ofthe definite integral 
 2
0
dx
x
tan , is
(A) 
2 (B*)
2

(C) 
2
2 (D)
2
2

[Sol. I = 
 2
0
dx
x
tan ....(1) ; I = 
 2
0
dx
x
cot ....(2)
adding (1) and (2), we get
2I =  



2
0
dx
x
cot
x
tan = 


2
0
dx
x
2
sin
x
cos
x
sin
2 = 




2
0
2
dx
)
x
cos
x
(sin
1
x
cos
x
sin
2
= 
 
1
1
2
t
1
dt
2 =  
1
0
2
t
1
dt
2
2 = 
2 (where sin x – cos x = t)
 I =
2

Ans. ]
Q.7629/inde Evaluate the integral:  dx
x
)
x
6
(
n 2
l
(A)
3
2
)]
x
6
(
n
[
8
1
l + C (B*) )]
x
6
(
n
[
4
1 2
2
l + C
(C) )]
x
6
(
n
[
2
1 2
l + C (D)
4
2
)]
x
6
(
n
[
16
1
l + C
[Hint: ln(6x2) = t ]
Q.77123/def 













6
5
6
2
2
d
)
sin
1
(
2
1
)
sin
3
(
2
1
(A)  – 3 (B*)  (C)  – 3
2 (D)  + 3
Q.78125/def Let l = 


x
2
x
x t
dt
Lim and m = 

x
Lim 
x
1
dt
t
n
x
n
x
1
l
l
then the correct statement is
(A*) l m = l (B) l m= m (C) l = m (D) l > m

[Hint: l = 

x
Lim ln 2x – ln x = ln 2 ; m =
x
n
x
t
n
x
1
l
l

= 

x
Lim
x
n
x
1
·
x
x
n
l
l

= 

x
Lim
x
n
1
x
n
l
l
 = 1
Hence l × m = ln 2 · 1 = ln 2 = l ]
Q.79126/def If f (x) = e–x + 2 e–2x + 3 e– 3x +...... +  , then 
3
n
2
n
dx
)
x
(
f
l
l
=
(A) 1 (B*)
2
1
(C)
3
1
(D) ln 2
[Sol. 
3
n
2
n
l
l
(e–x + 2 e–2x + 3 e– 3x +...... +  ) dx = –   3
n
2
n
x
4
x
3
x
2
x
......
e
e
e
e
l
l



 



= 























 .......
2
1
2
1
2
1
.......
3
1
3
1
3
1
3
2
3
2
=
2
1
2
1
1
3
/
1
1
3
/
1
2
/
1
1
2
/
1






Ans: ]
Q.80127/def If I = n x
(sin )
/
0
2

 dx then n x x
(sin cos )
/
/





4
4
dx =
(A*)
I
2
(B)
I
4
(C)
I
2
(D) I
[Hint: I1 =




/
/
4
4
ln (sin x + cos x) dx =




/
/
4
4
ln (cos x  sin x) dx (usingking)
 2 I1 =




/
/
4
4
ln cos 2x dx = 2
0
4
/
 ln (cos 2x) =
0
2
/
 ln (cos t) dt where 2x = t
0
2
/
 ln (sin t) dt = I  I1 = I/2 ]
Q.81129/def The value of  
 



















1
0
n
1
k
n
1
r
dx
k
x
1
)
r
x
( equals
(A) n (B) n ! (C) (n + 1) ! (D*) n · n !
[Hint: The givenintegrand is perfect differentialcoeff. of 


n
1
r
)
r
x
(
 I =
1
0
n
1
r
)
r
x
( 








= (n + 1)! – n! = n · n! ]

Q.8233/inde  

x
sin
x
sin
x
cos
x
cos
4
2
5
3
dx
(A) sinx  6 tan1 (sinx) + c (B) sin x  2 sin1 x + c
(C*) sinx  2 (sinx)1  6 tan1 (sin x) + c (D) sinx  2 (sinx)1 + 5 tan1 (sinx) + c
[Hint: sin x = t ; I =
  
 
 


2
2
2
2
t
1
t
t
2
t
1
dt =  


)
y
1
(
y
)
2
y
(
)
1
y
(
dy = 1 +
)
1
y
(
y
)
y
2
1
(
2


; y = t2
= 1 + 6 







1
y
1
y
3
1
= 









 2
2
t
1
6
t
2
1 dt ]
Q.83130/def
0
3

1
4 4
4 4
2
2
x x
x x
 
  







 dx =
(A) ln
5
2
3
2
 (B)ln
5
2
3
2
 (C*) ln
5
2
5
2
 (D) none
[Hint: I =  


3
0
dx
)
|
2
x
|
|
2
x
|
1
( =  
3
0
dx
2
x
dx
+  
3
0
dx
|
2
x
| ]
Q.84132/def The value ofthe function f(x) = 1+ x+
1
x
 (ln2t + 2lnt) dt where f (x) vanishes is :
(A) e1 (B) 0 (C) 2 e1 (D*) 1 + 2 e1
[Hint: f (x) = 1 + ln2 x + 2 ln x = 0  (1 + ln x)2 = 0  x =
1
e
Hence f 





e
1
= 1 +
e
1
+  
e
1
1
2
dt
)
nt
2
t
n
( l
l =
e
1
1
2
t
ln
t
e
1
1 



 =
e
1
e
1
1 
 = 1+ 2e–1
 [D]
Q.85134/def
Limit
n  
1
1
1 2 3 3 1
n
n
n
n
n
n
n
n
n n






 
 








.......
( )
has the value equalto
(A) 2 2 (B) 2 2  1 (C*) 2 (D) 4
[Hint : Tr =
1
n
n
n r

 S =
1
n 

 
3
n
3
0
r r
n
n
=
1
1
0
3

 x
dx ]
Q.86136/def Let a function h(x) be defined as h(x) = 0, for all x  0. Also 



dx
)
x
(
·
)
x
( f
h = f (0), for every
function f (x). Thenthe valueofthe definite integral 



dx
x
sin
·
)
x
(
'
h , is
(A) equalto zero (B) equalto 1 (C*) equalto – 1 (D)nonexistent
[Sol. I =  




dx
x
sin
·
)
x
(
'
I
II
h = 


)
x
(
h
·
x
sin – 



dx
)
x
(
h
·
x
cos = 0 – cos 0 = – 1  (A)
note that here cos x = f (x) ]

Q.87137/def
0
4
/
 (tann x + tann  2 x)d(x  [x]) is : ( [. ] denotes greatest integer function)
(A*)
1
1
n 
(B)
1
2
n 
(C)
2
1
n 
(D) noneofthese
Q.88139/def



 









1
1
0
0
dx
)
x
1
(
Lim is equal to
(A) 2 ln 2 (B*)
e
4
(C) ln
e
4
(D) 4
[Sol.



 









1
1
0
0
dx
)
x
1
(
Lim =




 












1
1
0
1
0 1
)
x
1
(
Lim =




 










1
1
0 1
1
2
Lim (1 form)
=


















 1
1
1
2
1
Lim
1
0
e =

















 )
1
(
2
2
Lim
1
0
e =














1
)
1
2
(
2
Lim
0
e = e2 ln 2 – 1 =






e
4
n
e
l
=
e
4
]
Q.8934/inde Which one ofthe following is TRUE.
(A) C
|
x
|
n
x
x
dx
.
x 

 l (B*) Cx
|
x
|
n
x
x
dx
.
x 

 l
(C) C
x
tan
dx
x
cos
.
x
cos
1


 (D) C
x
dx
x
cos
.
x
cos
1



[Hint: )
C
|
x
|
n
(
x
x
dx
.
x 

 l = x ln | x | + Cx ]
Q.90140/def
0

 x2n + 1·e x
 2
dx is equal to (n  N).
(A) n ! (B) 2 (n !) (C*)
n !
2
(D)
2
)!
1
n
( 
[Sol. I = 


0
x
n
2
dx
e
x
.
)
x
(
2
put x2 = t  x dx = – dt/2
= 


0
t
n
dt
e
t
2
1
= 








 





0
t
1
n
0
t
n
dt
e
t
n
e
t
2
1
=








 



0
t
1
n
dt
e
t
n
0
2
1
Hence I =
2
!
n
]

Q.91141/def The true set of values of'a' for which the inequality 
0
a
(32x  2. 3x) dx  0 is true is:
(A) [0, 1] (B) ( , 1] (C) [0, ) (D*) ( ,  1]  [0, )
[Sol.  



0
a
x
x
0
dx
)
2
3
(
3 put 3–x = t 3–x ln3 dx = –dt




a
3
1
0
dt
)
2
t
(
3
ln  0
t
2
2
t
a
3
1
2






0
2
2
1
3
.
2
2
3 a
a
2

















 

3–2a – 4×3–a + 3 > 0
(3–a – 3) (3–a – 1) > 0
3–a > 31
 a < 1
or 3–a < 30
 a > 0
Hence )
,
0
[
)
1
,
(
a 



 ]
Q.92142/def If (2, 3) then number ofsolution of the equation 

0
cos(x+ 2) dx = sin is :
(A) 1 (B*) 2 (C) 3 (D) 4.
[Sol. 

 0
2
)
x
sin( = sin
sin(2 + ) – sin2 = sin
2 cos(2 +/2) sin/2 = sin
now proceed and get
2 ,
  
1 1 8
2

 2 solutions ]
Q.93143/def If x · sin x = 
2
x
0
dt
)
t
(
f where f is continuous functions then the value of f(4) is
(A*)
2

(B) 1 (C)
2
1
(D) cannot be determined
[Hint: differentiate w.r.t. x, 2x f (x2) = sin x + x cos x
put x = 2, 4· f (4) = 0 + 2  f (4) =
2

; when x = – 2, (–4) · f (4) = – 2, f (4) =
2

]
Q.9436/inde  


dx
)
1
x
4
x
(
)
1
x
2
(
2
/
3
2
(A) C
)
1
x
4
x
(
x
2
/
1
2
3



(B*) C
)
1
x
4
x
(
x
2
/
1
2



(C) C
)
1
x
4
x
(
x
2
/
1
2
2



(D) C
)
1
x
4
x
(
1
2
/
1
2




[Hint:  


dx
)
1
x
4
x
(
1
x
2
2
/
3
2
= 









dx
x
1
x
4
1
x
1
x
2
2
/
3
2
3
= 








 

dx
x
1
x
4
1
x
x
2
2
/
3
2
3
2
now put
2
2
t
1
x
4
x
1


 ]
Q.95144/def Ifthe value ofthe integral ex2
1
2
 dxis , thenthe value of nx
e
e4
 dxis :
(A) e4  e  (B*) 2 e4  e  (C) 2 (e4  e)  (D) 2 e4 – 1 – 
[Hint : Put ln x = t2  x =et2
 dx = 2t et2
dt.
Hence I =
1
2
 t. 2 t et2
dt.
Now I.B.P. takingt as the first functionand 2 tet2
as the second function ]
Q.96145/def








 2
x
1
x
2
1
tan
dx
d
2
1
3
0
equals
(A*)
3

(B)
6

 (C)
2

(D)
4

[Hint: tan–1 





 2
x
1
x
2
= 


1
x
if
x
tan
2
1
x
0
if
x
tan
2
1
1







]
Q.97146/def Let A=
0
1

e d t
t
t
1 
then 




a
1
a
t
1
a
t
dt
e
has the value
(A)Aea (B*) Aea (C)  aea (D)Aea
[Hint: I = 




a
1
a
t
dt
1
a
t
e
put t = a–1+ y(so that lower limit becomes zero)
I =  


1
0
y
a
1
dy
2
y
e
(nowusingking)
I =  




1
0
y
1
a
1
dy
2
y
1
e
=  
 
1
0
y
a
dy
y
1
e
e = – e–a A (B) ]
Q.98148/def
sin
/
2
0
2


 sin d is equal to :
(A) 0 (B*) /4 (C) /2 (D) 

[Hint: I = sin
/
2
0
2


 . cos  d 2 I = sin
/
2
0
2


 (cos  + sin ) d
=  
1
2
0
2
 
 sin cos
/
 

. (cos  + sin ) d Put sin  cos  = t ]
Q.9937/inde dx
4
x
2
x
4
2
 

is equal to
(A) C
x
2
2
x
tan
2
1 2
1



(B) C
)
2
x
(
tan
2
1 2
1



(C) C
2
x
x
2
tan
2
1
2
1



(D*) C
x
2
2
x
tan
2
1 2
1



Q.100150/def If  + 2 x e x
2
0
1
2

 dx = e x


2
0
1
dx then the value of  is
(A*) e1 (B) e (C) 1/2e (D) cannot be determined
[Hint:  +
0
1
 x. 2 xe x
 2
d x =
0
1
 e x
 2
d x
 +  
 
xe x2
0
1

0
1
 e x
 2
d x =
0
1
 e x
 2
d x  =
1
e
]
Q.101151/defAquadratic polynomialP(x) satisfies the conditions, P(0) = P(1) = 0 &
0
1
 P(x) dx= 1.The leading
coefficient ofthequadratic polynomialis :
(A) 6 (B*) 6 (C) 2 (D) 3
[Hint : Take P(x) =Ax (x  1) and now compute Ausing , A
0
1
 x (x  1) dx = 1 ]
Q.102152/def Whichone ofthe following functions is not continuous on (0,)?
(A) f(x)= cotx (B) g(x) = 
x
0
dt
t
1
sin
t
(C) h (x) =











x
4
3
x
9
2
sin
2
4
3
x
0
1
(D*) l (x) =














x
2
,
)
x
sin(
2
2
x
0
,
x
sin
x
[Sol. g (x) = dt
t
1
sin
t
x
0

g (x) = x sin(1/x) which is diff  cont. in(0, )
l (x) = 











x
2
/
x
sin
2
/
2
/
x
0
x
sin
x
obvious discontinuityat x = /2  (D) ]

Q.103153/def If f (x) = 


0
2
2
t
sin
x
tan
1
dt
t
sin
t
for 0 < x <
2

(A) f (0+) = –  (B)
8
4
f
2







 
(C*) fis continuous and differentiablein 




 
2
,
0
(D) fis continuous but not differentiable in 




 
2
,
0
[Sol. f (x) = dt
t
sin
x
tan
1
t
sin
t
0
2
2



Using king and add.
f (x) = dt
t
sin
x
tan
1
t
sin
2 0
2
2




= dt
)
t
cos
1
(
x
tan
1
t
sin
2
/
0
2
2





= dt
t
cos
x
tan
x
sec
t
sin
2
/
0
2
2
2



 =  

1
0
2
2
2
y
.
x
tan
x
sec
dy
=  

1
0
2
2
y
x
ec
cos
dy
x
tan
=
x
tan
x
)
x
(sin
sin
x
tan
ecx
cos
y
sin
x
tan
1
1
0
1 









 

]
Q.10441/inde Consider f(x) =
x
x
2
3
1
; g(t) = f t dt
( )
 . Ifg(1) = 0 theng(x) equals
(A)
1
3
1 3
n x
( )
 (B*)
1
3
1
2
3
n
x






 (C)
1
2
1
3
3
n
x






 (D)
1
3
1
3
3
n
x







Q.105158/def The value ofthe definite integral, 
100
0
x
dx
e
x
2 is equalto
(A)
2
1
(1 – e–10) (B) 2(1 – e–10) (C)
2
1
(e–10 – 1) (D*)
2
1





   4
10
e
1
[Hint: put x2 = t; x dx =
2
dt
I =
2
1

4
10
0
t
e
dt
= –
2
1

4
10
0
t
e =
2
1





   4
10
e
1 Ans. ]
Ans. ]

Q.106159/def
0

 [2 ex] dx where [x] denotes the greatest integer function is
(A) 0 (B*) ln 2 (C) e2 (D) 2/e
[Hint: for 0 < x < ln 2, [2ex] = 1, otherwise zero  I =
0
2
n
 dx +
n2

 0 dx = ln 2
alternatively, put ex = t ; – x = ln t ; dx = dt
t
1

1
0
dt
t
1
]
t
2
[ – 
1
0
t
dt
]
t
2
[ ; 
2
1
0
dt
0 + 
2
1
2
1
t
dt
= 1
2
1
t
n
l = [1] – ln
2
1
= ln 2 Ans.]
Ans.]
Q.107160/def The value of 

1
1
|
x
|
dx
is
(A)
2
1
(B) 2 (C*) 4 (D) undefined
[Sol. 
1
0
x
dx
2 =
1
0
1
2
1
1
2
1
x














= 4  1
0
x = 4  (C) ]
Q.108161/def
x n
x
dx
l 1
2
0
1







 =
(A*)
3
4
1 2
3
2







ln (B)
3
2
7
2
3
2
 ln (C)
3
4
1
2
1
54
 ln (D)
1
2
27
2
3
4
ln 
[Sol. I =  




 
1
0
dx
2
2
x
n
x l =  

1
0
dx
)
2
n
)
2
x
(
n
(
x l
l
= 
 

1
0
1
0
dx
x
2
n
dx
)
2
x
(
n
x l
l   






1
0
2
1
0
2
2
2
n
dx
2
x
x
2
x
.
)
2
x
(
n
l
l
=  




1
0
2
2
2
n
dx
2
x
4
4
x
3
n
2
1 l
l   









1
0
dx
2
x
4
)
2
x
(
2
3
n
2
1
l now proceed]
Q.10942/inde The evaluation of
p x q x
x x
dx
p q q
p q p q
  
 

 
z 2 1 1
2 2
2 1
is
(A) –
x
x
C
p
p q



1
(B)
x
x
C
q
p q



1
(C*) 



x
x
C
q
p q
1
(D)
x
x
C
p
p q



1

[Sol.
p x qx
x
p q q
p q
  



z 2 1 1
2
1
( )
dx =
p x qx
x x
dx
p q
p q
  



z 1 1
2
( )
taking xq as x2q common fromDenominator and take it in Nr ]
Q.110162/def
x x
x x
3
2
1
1
1
2 1
 
 


| |
| |
dx = a ln 2 + b then :
(A) a = 2 ; b = 1 (B*) a = 2 ; b = 0 (C) a = 3 ; b =  2 (D) a = 4 ; b =  1
[Hint: I = 



1
1
2
3
1
x
2
x
x
dx +
 
x
x




1
1
2
1
1
dx  2
dx
x
1
0
1

 = 2 ln 2 ]
odd  vanishes even ]
Q.111163/def
a
b
 [x] dx +
a
b
 [x] dxwhere [. ] denotes greatest integer function is equalto :
(A) a + b (B) b  a (C*) a  b (D)
a b

2
[Hint: [x] + [ x] =
0
1
if x I
if x I

 


  I =
a
b
  dx = a  b ]
Q.112166/def If
0
2
 375 x5 (1 + x2) 4 dx = 2n then the value of n is :
(A) 4 (B*) 5 (C) 6 (D) 7
[Sol.
n
2
0
4
2
5
2
dx
)
x
1
(
x
375


 ;
n
2
0
4
2
8
5
2
dx
)
x
1
(
x
x
375


  ;
n
2
0
4
2
3
2
dx
)
x
1
(
x
375


 

simplifying
n
2
0
3
2
2
2
x
1
x
6
375














n
2
5
.
5
.
5
4
.
4
.
4
6
375
  2n = 32  n = 5 ]
Q.113167/def  


2
/
1
0
2
x
1
x
1
n
x
1
1
 dx is equal to :
(A*)
3
1
n
4
1 2
 (B)
2
1
ln2 3 (C) 
4
1
ln2 3 (D) cannot be evaluated.
[Hint: Put ln (1 + x)  ln (1  x) = t 
dx
x
1 2

=
1
2
dt
I =
1
2 0
3
n
 t dt =
1
4
ln2 3 =
1
4
ln2
1
3
]

Q.11443/inde If  
 dx
e
)
5
x
2
x
( x
3
2
3
= e3x (Ax3 + Bx2 + Cx+ D) then the statement which is incorrect is
(A) C + 3D = 5 (B) A + B + 2/3 = 0
(C*) C + 2B = 0 (D) A+ B + C = 0
[Hint: A =
3
1
, B = – 1, C =
3
2
; D =
9
13
]
Q.115169/def Given 



2
/
0
x
cos
x
sin
1
dx
= ln 2, then the value of the def. integral. 



2
/
0
x
cos
x
sin
1
x
sin
dx is
equalto
(A)
2
1
ln2 (B)
2

 ln 2 (C*)
4

–
2
1
ln 2 (D)
2

+ ln 2
[Hint: I =
cos
sin cos
/
x
x x
1
0
2
 


 2 I =
sin cos
sin cos
/
x x
x x
  
 

1 1
1
0
2

dx
 2 I =

2
 ln 2  I =

4
1
2
 ln 2 ]
Q.116171/def Afunction f satisfying f (sinx) = cos2 x for allx and f(1) = 1 is :
(A) f(x) = x +
3
1
3
x3
 (B) f(x) =
3
2
3
x3

(C*) f(x) = x 
3
1
3
x3
 (D) f(x) = x 
3
1
3
x3

[Hint: Put sin x = t  f (t) = 1  t2. Now integrate ]
Q.117172/def For 0 < x <

2
,
1 2
3 2
/
/
 ln (ecos x). d (sinx) is equal to :
(A*)
12

(B)
6

(C)    
 
1
sin
3
sin
1
3
4
1


 (D)    
 
1
sin
3
sin
1
3
4
1



[Hint : I =


/
/
6
3
 cos x cos x dx =


/
/
6
3
 cos2 x dx =


/
/
6
3
 sin2 x dx  2 I =


/
/
6
3
 dx =

6
 I =

12
]
Q.118173/def  



0
2
x
sin
1
x
cos
x
dx is equal to :
(A) 2 (B)  (2 + ) (C) zero (D*) 2 
[Hint: Take x as the first and
 
cos
sin
x
x
1
2

as the second function. I B P]

Q.11944/inde   dx
x
x
x
e x

(A*) 2  
1
x
x
e x

 + C (B)  
1
x
2
x
e x


(C)   C
x
x
e x

 (D)   C
1
x
x
e x



[Sol.   dx
x
x
x
e x
 ; put x = t2 ; dx = 2t dt
=   dt
)
t
t
(
e 2
t
= et (At2 + Bt + C) (Let)
Diffrentiate boththe sides
et (t2 + t) = et (2At + B) + (At2 + Bt + C) et
Oncomparingcoefficient we get
A = 1 ; B = – 1 ; C = 1 ]
Q.120174/def
dx
x x
cos sin
/
6 6
0
2



is equal to :
(A) zero (B*)  (C) /2 (D) 2 
[Hint: I =
dx
x x
1 3 2 2
0
2

 sin cos
/

=
dx
x
1 2
3
4
2
0
2

 sin
/

= 2
dt
t
4 3 2
0 
 sin

where 2x = t
= 4
dt
t
4 3 2
0
2

 sin
/

etc. ]
Q.121178/def The true solution set ofthe inequality, 




x
0
2
2
x
6
x
5 dz > x
0

 sin2 x dxis :
(A) R (B) (1, 6) (C) ( 6, 1) (D*) (2, 3)
[Hint: 5 6 2
x x
  +
2
x

> x

2
 5x  6  x2 > 0 ]
Q.122179/def If  
1
0
2
x
1
x
n

dx = k
0

 ln (1 + cosx) dx then the value of k is :
(A) 2 (B*) 1/2 (C)  2 (D) 1/2
[Hint: Put x = sin   L H S =
0
2
/
 ln (sin ) d = k
0

 ln (1  cos x)
= 2 k
0
2
/
 ln (sin x) dx  k = 1/2 ]
Q.123182/def Let a, b and c be positive constants. The value of 'a' in terms of 'c' if the value of integral




1
0
5
b
3
3
1
b
dx
)
bx
a
acx
( is independent of b equals
(A*)
2
c
3
(B)
3
c
2
(C)
3
c
(D)
c
2
3

[Sol. I = 



1
0
5
b
3
3
1
b
dx
)
bx
a
acx
( ; ac·
1
0
6
b
3
3
2
b
6
b
3
bx
a
2
b
x








=
)
2
b
(
3
b
a
2
b
ac 3



I(b) =
)
2
b
(
3
1

[3ac + a3b] 
)
2
b
(
3
a
c
3
b
a 2
3








Ifthis is independent of b, then 2
a
c
3
2

Alternatively: Ifthe integralis independent ofb, then I'(b) = 0
 I'(b) = 2
)
2
b
(
ac


+ 
















2
b
2
2
b
·
3
a
D
3
= 2
)
2
b
(
ac


+ 







 2
3
)
2
b
(
2
3
a
= 2
)
2
b
(
1
 







 ac
3
a
2 3
=
3
a
2 3
= ac  a =
2
c
3
Ans.]
Ans.]
Q.12451/inde 




 d
)
tan
(sec
sec 2
2
(A) C
)]
tan
(sec
tan
2
[
2
)
tan
(sec









(B) C
)]
tan
(sec
tan
4
2
[
3
)
tan
(sec









(C*) C
)]
tan
(sec
tan
2
[
3
)
tan
(sec









(D) C
)]
tan
(sec
tan
2
[
2
)
tan
(sec
3









[Hint: put sec  + tan  = t]
Q.125184/def  

2
1
4
2
1
x
1
x
dx is equal to:
(A)
1
2
tan1
2 (B*)
1
2
cot1 2 (C)
1
2
tan1 1
2
(D)
1
2
tan1 2
Q.126186/def 1
x
x
Limit

1
x
x
x


x
x1
f(t) dt is equal to :
(A)
 
f x
x
1
1
(B*) x1 f(x1) (C) f(x1) (D) does not exist
[Sol.





 


x
x
x
dt
)
t
(
f
Limit
1
x
x
x
x
1
1
=
1
2
x
x x
x
.
)
x
(
f
Limit
1

(using Lopital's rule) = x1 f(x1)  (B) ]

Q.127187/def Which ofthe following statements could be true if, f(x) = x1/3.
I II III IV
f(x) =
28
9
x7/3 + 9 f (x) =
28
9
x7/3  2 f (x) =
4
3
x4/3 + 6 f(x) =
4
3
x4/3  4
(A) I only (B) III only (C) II & IV only (D*) I & III only
[Sol. f   (x) = x1/3  f  (x) = 3/4 x4/3 + C1 .....(1)
f (x) = 2
1
3
/
7
C
x
C
x
3
7
.
4
3

 = 2
1
3
/
7
C
x
C
x
4
7

  I if C1 = 0 ]
Q.128191/def The value ofthe definite integral
0
2
/
 sinx sin2x sin3x dx is equalto :
(A)
1
3
(B) 
2
3
(C) 
1
3
(D*)
1
6
[Hint: Use
0
a
 f(x) =
0
a
 (a  x) dx and add two integrals ]
Q.12952/inde
dx
x
1
x
1
cos
x
1
sec
)
x
1
(
e
2
2
1
2
2
1
2
x
1
tan
 























 




(x > 0)
(A) C
x
tan
.
e 1
x
1
tan



(B)
  C
2
x
tan
.
e
2
1
x
1
tan



(C*) C
x
1
sec
.
e
2
2
1
x
1
tan












 


(D) C
x
1
ec
cos
.
e
2
2
1
x
1
tan












 


[Hint: note that sec–1 2
x
1 = tan–1x ; cos–1










2
2
x
1
x
1
= 2 tan–1x for x > 0
I =  dx
x
tan
2
)
x
(tan
x
1
e 1
2
1
2
x
tan 1






put tan–1x = t
= dt
)
t
2
t
(
e 2
t

 = et . t2 =  2
1
x
1
tan
x
tan
e 

+ C ]
Q.130193/def Number of positive solution of the equation,  
 
t t
x


2
0
dt = 2 (x  1) where { } denotes the
fractionalpart functionis :
(A) one (B*) two (C) three (D) morethanthree
[Hint: consider 0 < x < 1 ; 1  x < 2 ; 2  x < 3 etc. Ans. x = 1 and 5/2 ]

Q.131194/def Iff(x) = cos(tan–1x) then the value of the integral dx
)
x
(
'
'
f
x
1
0
 is
(A)
2
2
3
(B)
2
2
3
(C) 1 (D*)
2
2
3
1
[Sol. f(x) = cos (tan–1x)
f  (x) = 2
1
x
1
)
x
sin(tan



I = dx
)
x
(
'
'
f
x
1
0
 =  dx
)
x
(
'
f
)
x
(
'
f
x
1
0
1
0 

= f  (1) – 1
0
)
x
(
f = f  (1) – [ f (1) – f (0) ] = f  (1) – f (1) + f (0)
f (0) = 1 ; f  (1) = –
2
2
1
; f (1) =
2
1
]
Q.1322/inde If  
2
x
sin
1 dx =Asin 




 

4
4
x
then value ofA is:
(A) 2 2 (B) 2 (C)
1
2
(D*) 4 2

Q.133201/def For Un =
0
1
 xn (2  x)n dx; Vn =
0
1
 xn (1  x)n dx n  N, which of the following statement(s)
is/are ture?
(A) Un = 2n Vn (B) Un = 2 n Vn (C*) Un = 22n Vn (D) Un = 2  2n Vn
[Sol. Given Un =  
1
0
n
n
dx
)
x
2
(
.
x ; Vn =  
1
0
n
n
dx
)
x
1
(
.
x
put x = 2t  dx = 2dt
 Un =  
2
/
1
0
n
n
n
n
dt
)
t
1
(
2
t
.
2
2 ....(1)
Now Vn =  
2
/
1
0
n
n
dx
)
x
1
(
x
2 (Using Queen) .....(2)
From (1) and (2)
Un = 22n. Vn  (C) ]
Q.13454/inde
 






 




x
1
x
tan
)
1
x
3
x
(
dx
)
1
x
(
2
1
2
4
2
= ln | f (x) | + C then f (x) is
(A) ln 






x
1
x (B*) tan–1 






x
1
x (C) cot–1 






x
1
x (D) ln 















x
1
x
tan 1
[Hint:
 dx
x
1
x
tan
3
x
1
x
x
1
1
1
2
2
2

















;
 dx
x
1
x
tan
1
x
1
x
dx
x
1
1
1
2
2































]
Q.135202/def Let f(x) be integrable over (a, b) , b > a > 0. If I1 =


/
/
6
3
 f (tan  + cot ). sec2  d &
I2 =


/
/
6
3
 f (tan  + cot ). cosec2  d , thenthe ratio
I
I
1
2
:
(A*) is a positive integer (B) isa negative integer
(C) is anirrationalnumber (D) cannot be determined.
[Hint: Using king in I1 , I1 =


/
/
6
3
 f(tan  + cot ). cosec2  d = I2 
I
I
1
2
= 1 ]

Q.136203/def f(x) =
cos
sin
x
x
 (1  t + 2 t3) d t has in [ 0, 2  ]
(A) a maximumat

4
& aminimumat
3
4

(B*) amaximumat
3
4

& aminimumat
7
4

(C) a maximumat
5
4

&a minimumat
7
4

(D) neither amaxima nor minima
[Hint : f (x) = (1. cosx  sinx cosx + 2 sin3 x. cos x)  (1. sinx + cosx sinx  2 cos3 x sin x)
= cosx + sinx – 2 sinx · cosx + 2 sinx · cosx = cosx + sinx ]
Q.137206/def Let S (x) =
x
x
2
3
 ln t d t (x > 0) and H(x) =

S x
x
( )
. Then H(x) is :
(A) continuousbut not derivable inits domain
(B*) derivable andcontinuous inits domain
(C) neither derivable nor continuous inits domain
(D) derivablebut not continuous inits domain.
[Hint: S  (x) = ln x3. 3x2  ln x2. 2x = 9x2 ln x  4x ln x
= x l n x (9 x  4). Hence

S x
x
( )
= ln x (9 x  4).
Now it is obvious that

S x
x
( )
iscontinuous and derivable inits domain. ]
Q.138211/def Number ofsolution ofthe equation
d
dx

x
sin
x
cos
dt
t
1 2

= 2 2 in [0, ] is
(A) 4 (B) 3 (C*) 2 (D) 0
[Hint: LHS = secx + cosecx =2 2  x =

4
and
11
12

]
Q.13955/inde Let f (x) =
x
cos
1
x
sin
2 2

+
x
sin
1
)
1
x
sin
2
(
x
cos


then
 
  dx
)
x
(
'
f
)
x
(
f
ex
(where c is the constant ofintegeration)
(A*) ex tanx + c (B) excotx + c (C) ex cosec2x + c (D) exsec2x + c
[Hint:
x
sin
1
)
x
sin
2
1
(
x
cos


–
x
cos
x
sin
x
cos 2
2

=
)
x
sin
1
(
x
cos
)
x
sin
x
)(cos
x
sin
1
(
)
x
sin
2
1
(
x
cos 2
2
2





=
)
x
sin
1
(
x
cos
x
sin
x
cos
x
sin 3
2



=
)
x
sin
1
(
x
cos
)
x
sin
1
(
x
sin
x
cos
x
sin 2
2



=
x
cos
x
sin
)
x
sin
1
(
x
sin 2


= tan x ]

Q.140212/def The value ofx that maximises the value ofthe integral t t dt
x
x
( )
5
3


 is
(A) 2 (B) 0 (C*) 1 (D) none
[Sol. F (x) = dx
)
t
5
(
t
3
x
x



F  (x) = (x + 3) (2 – x) – x (5 – x) = 6 – x – x2 – 5x + x2 = 6 – 6x = 0  x = 1
F  (x) = – 6  x =1 is maxima  (C) ]
Q.141213/def For a sufficiently large value of n the sum of the square roots of the first n positive integers
i.e. 1 2 3
   
...................... n is approximatelyequalto
(A)
1
3
3 2
n /
(B*)
2
3
3 2
n /
(C)
1
3
1 3
n /
(D)
2
3
1 3
n /
[Hint: Limit
n
n
n n

   
1 2 3 ....................
= x dx
0
1
 =
2
3
 Sn =
2
3
3 2
n /
]
Q.142216/def The value of 
2
0
2
)
x
1
(
dx
is
(A) –2 (B) 0 (C) 15 (D*) indeterminate
[Sol.  
2
0
2
)
x
1
(
dx
= 
 




2
1
2
1
0
2
)
x
1
(
dx
)
x
1
(
dx
=
2
1
1
0 x
1
1
x
1
1











= ( – 1) + (–1) – (– )  indeterminant
Note that the shaded area is divergent ]
Q.143217/def If 








8
/
0
a
0
d
2
sin
tan
2
x
a
x
dx
, then the value of 'a' is equal to (a > 0)
(A)
4
3
(B)
4

(C)
4
3
(D*)
16
9
[Sol.  
 

a
0
dx
x
a
x
a
1
= 





d
sec
.
tan
2
tan
2 2
8
/
0
(sin2 =



2
tan
1
tan
2
)
=  
a
0
2
/
3
2
/
3
x
)
a
x
(
3
2
.
a
1
 =
8
/
0
tan

 =    
1
2
a
a
)
a
2
(
a
3
2 2
/
3
2
/
3
2
/
3




=   1
2
1
2
a
2
.
a
3
2 2
/
3


  1
a
3
4
 
16
9
a  ]

Q.14459/inde The value ofthe integral
 
 

dx
1
x
)
x
2
2
(
n
sin l
is
(A*) – cos ln (2x + 2) + C (B) ln 





1
x
2
sin + C
(C) cos 





1
x
2
+ C (D) sin 





1
x
2
+ C
[Hint: ln 2(1 + x) = t ; dx
)
x
1
(
1

= dt]
Q.145218/def If f(x) =Asin 




 
2
x
+ B , f 





2
1
= 2 and 
1
0
f(x) dx =

A
2 , Then the constantsAand B are
respectively.
(A)
2
&
2


(B)


3
&
2
(C)


4
&
0 (D*) 0
&
4

Q.146219/def Let I1 = 


2
0
x
dx
)
x
sin(
e
2
; I2 = 


2
0
x
dx
e
2
; I3 = 



2
0
x
dx
)
x
1
(
e
2
and considerthe statements
I I1 < I2 II I2 < I3 III I1 = I3
Whichofthefollowing is(are) true?
(A) I only (B) II only
(C) Neither I nor II nor III (D*) BothI and II
[Sol. since 0 < sin x < 1 and 1 + x > 1 in (0, /2)
hence I3 > I2 > I1
 A and B are correct  (D) ]
Q.147220/def Let f (x) =
x
x
sin
, then 









2
0
dx
x
2
f
)
x
(
f =
(A*) 

 0
dx
)
x
(
f
2
(B) 

0
dx
)
x
(
f (C) 


0
dx
)
x
(
f (D) 

 0
dx
)
x
(
f
1
[Hint: I = 









2
0
dx
x
2
x
x
cos
x
sin
= 



2
0
dx
)
x
2
(
x
x
2
sin
; put 2x = t
I = 



0
dt
)
t
(
t
t
sin
= 












 0
dt
)
t
(
t
sin
t
t
sin
1
= 

 0
dt
t
t
sin
2
]
Q.148222/def Let u =  

1
0
2
dx
1
x
)
1
x
(
n
l
and v = 
 2
0
dx
)
x
2
(sin
n
l then
(A) u = 4v (B*) 4u + v = 0 (C) u + 4v = 0 (D) 2u + v = 0

[Sol. u =  

1
0
2
dx
1
x
)
1
x
(
n
l
put x = tan 
= 




4
0
d
)
tan
1
(
n
l = 













4
0
d
tan
1
tan
1
1
n
l = 




4
0
d
tan
1
2
n
l =
4

ln 2 – u
 u =
8

ln 2  4u =
2

ln 2 ....(1)
again v = 
 2
0
dx
)
x
2
(sin
n
l (put 2x = t) ; v = 

0
dt
)
t
(sin
n
2
1
l = 
 2
0
dx
)
t
(sin
n
l
v = –
2

ln 2 ....(2)
(1) + (2)  4u + v = 0  (C) ]
Q.149223/def If   



 

d
.
cos
1
·sin
x
sin
x
2
16
/
2
x
2
f then the value of f ' 




 
2
, is
(A*)  (B) –  (C) 2 (D) 0
[Sol.   



 

d
.
cos
1
sin
x
sin
x
2
16
/
2
x
2
f ; f ' (x) = x
cos
d
cos
1
sin
0
x
2
·
x
cos
1
x
sin
x
sin
2
2
x
16
2
2

























f ' 




 
2
=
1

Ans. ]
Q.150225/def The value of the definite integral, 
 2
0
dx
x
sin
x
5
sin
is
(A) 0 (B*)
2

(C)  (D) 2
[Sol. sin nx– sin(n – 2)x = 2 cos(n – 1)x sin x






 dx
x
sin
x
)
2
n
sin(
dx
)
1
n
cos(
2
dx
x
sin
nx
sin
 







2
0
2
0
2
0
dx
x
sin
x
3
sin
dx
x
4
cos
2
dx
x
sin
x
5
sin
=0 + 
 2
0
dx
x
sin
x
3
sin
= 
 2
0
dx =
2

Ans. ]
Select the correct alternatives : (More than one are correct)
Q.151502/def
dx
x
sgn
b
a
 = (where a, b R)
(A*) | b | – | a | (B) (b–a) sgn (b–a) (C*) b sgnb – a sgna (D) | a | – | b |

Q.152503/inde   x
cos
4
5
dx
=  tan1 





2
x
tan
m + C then :
(A*)  = 2/3 (B*) m = 3 (C)  = 1/3 (D) m = 2/3
Q.153503/def Which ofthe following are true ?
(A*) x f x
a
a
. (sin )

 dx =

2
. f x
a
a
(sin )
 
 dx (B*) f x
a
a
( )2

 dx = 2. f x
a
( )2
0
 dx
(C*)  
f x
n
cos2
0

 dx = n.  
f x
cos2
0

 dx (D*) f x c
b c
( )



0
dx = f x
c
b
( )
 dx
Q.154505/def The value of
 
2 3 3
1 2 2
2
2
0
1
x x
x x x
 
  
 ( )
dx is :
(A*)

4
+ 2 ln2 tan1 2 (B)

4
+ 2 ln2tan1
1
3
(C*) 2 ln2 cot1 3 (D*) 

4
+ ln4+ cot1 2
[Hint : Numerator = 2 (x2 + 2 x + 2)  (x + 1) ]
Q.155505/inde
x x
x
2 2
2
1



cos
cosec2 x dx is equal to :
(A) cot x  cot 1 x + c (B*) c  cot x + cot 1 x
(C*)  tan1 x 
cos
sec
ec x
x
+ c (D*) e
n x
 tan1
 cot x + c
where 'c' is constant ofintegration .
Q.156506/def Let f(x) =
sint
t
x
0
 dt (x > 0) then f(x) has :
(A*) Maxima if x = n where n = 1, 3, 5,.....
(B*) Minima if x = n where n = 2, 4, 6,......
(C) Maxima if x = n where n = 2, 4, 6,......
(D)Thefunctionismonotonic
[Hint: f (x) =
sinx
x
; f (x) =
x x x
x
cos sin

2
Now f (2 m) we have f (x) > 0 & f ((2 m  1) ) we have f (x) < 0 ]
Q.157508/def IfIn =
 
dx
x
n
1 2
0
1

 ;n  N, thenwhich of the following statements hold good ?
(A*) 2n In + 1 = 2 n + (2n  1) In (B*) I2 =

8
1
4

(C) I2 =

8
1
4
 (D) I3 =

16
5
48

[Hint: I.B.P. taking 1 as the 2nd and
1
1 2
( )
x n
as the 1st function ]

Q.158509/inde
1
1
1
1
2
x
n
x
x
dx



z  equals :
(A)
1
2
ln2
x
x


1
1
+ c (B*)
1
4
ln2
x
x


1
1
+ c (C)
1
2
ln2
x
x


1
1
+ c (D*)
1
4
ln2
x
x


1
1
+ c
[Hint: put ln (x  1)  ln (x + 1) = t ]
Q.159510/def IfAn =
0
2
/

sin ( )
sin
2 1
n x
x

d x ; Bn =
0
2
/

sin
sin
nx
x






2
d x ; for n  N , then :
(A*) An + 1 = An (B) Bn + 1 = Bn
(C) An + 1  An = Bn + 1 (D*) Bn + 1  Bn = An + 1
[Hint: Consider An + 1  An =
0
2
/

sin ( ) sin ( )
sin
2 1 2 1
n x n x
x
 
=
0
2
/
 2 cos 2 n x d x = 0
 An + 1 = An
Again consider Bn + 1  Bn =
0
2
/

sin ( ) sin
sin
2 1 2
2
n x x
x
 
d x =
0
2
/

sin ( )
sin
2 1
n x
x

dx = A
An + 1 ]
Q.160511/def
0


x
x x
( ) ( )
1 1 2
 
d x :
(A*)

4
(B)

2
(C*) is same as
0


dx
x x
( ) ( )
1 1 2
 
(D) cannot be evaluated
[Hint: Put x = 1/t and add the two integrals ]
Q.161512/inde 1 
 cscx dx equals
(A*) 2 sin1 sinx + c (B) 2 cos 1 cosx + c
(C) c  2 sin1 (1  2 sin x) (D*) cos1 (1  2 sin x) + c
[Hint: I =
   
 
1 1
1
 

sin sin
sin sin
x x
x x
=
 
cos
sin sin
x
x x
1
=
 
cos
sin
x
x
1
4
1
2
2
 
=
 



dt
t
1
2
2
2
]
Q.162512/def If f(x) =
0
2
/

n x
( )
sin
sin
1 2
2
 

d  , x  0 then :
(A*) f (t) =   
t  
1 1 (B*) f (t) =

2 1
t 
(C) f(x) cannot be determined (D) none ofthese.
[Sol. 







2
/
0
2
2
2
d
)
sin
x
1
(
sin
sin
dx
dI
= 





2
/
0
2
sin
x
1
d
dx
dI
MultiplyNr. and Dr. bysec2 and proceed ]

Q.163514/def If a, b, c  R and satisfy 3 a + 5 b + 15 c = 0 , the equation ax4 + bx2 + c = 0 has :
(A*) atleast one root in (1, 0) (B*) atleast one root in (0, 1)
(C*) atleast two roots in (1, 1) (D) no root in (1, 1)
[Hint:
0
1
 f(x) d x =
a b
5 3
 + c =
1
15
(3 a + 5 b + 15 c) = 0
 B Since f(x) is even  A  C ]
Q.164515/def Let u = 



0
2
4
1
x
7
x
dx
& v = 



0
2
4
2
1
x
7
x
dx
x
then :
(A) v > u (B*) 6 v =  (C*) 3u + 2v = 5/6 (D*) u + v = /3
[Hint: put x = 1/t in u or v  u = v. Now consider u + v ]
Q.165513/inde If  eu . sin2x dx can be found in terms ofknown functions of x then u can be :
(A*) x (B*) sin x (C*) cos x (D*) cos 2x
Q.166518/def Iff(x) =
n t
t
x
1
1 
 dt where x > 0 thenthe value(s) ofx satisfying the equation,
f(x) + f(1/x) = 2 is :
(A) 2 (B) e (C*) e 2 (D*) e2
[Hint: f(x) =
2
x
n2

= 2  C, D ]
Q.167519/def Apolynomialfunction f(x) satisfying the conditions f(x) = [f(x)]2 &
0
1
 f(x) dx =
12
19
canbe:
(A)
4
9
x
2
3
4
x2

 (B*)
4
9
x
2
3
4
x2

 (C)
4
x2
 x + 1 (D*)
4
x2
+ x + 1
Q.168520/def Acontinuous anddifferentiable function 'f' satisfies the condition ,
0
x
 f(t) dt = f2 (x)  1 for all real 'x'. Then :
(A*) 'f' is monotonic increasing  x R
(B) 'f' is monotonic decreasing  x R
(C) 'f' is non monotonic
(D*) the graphofy= f(x) is a straight line.
[Sol. Differentiating f(x) = 2 f(x). f  (x)  f (x) =
1
2
( f(x))  0)
Hence f(x) =
x
2
+ c. Put x = 0 ; f (0) = c ; but f2 (0) = 1
 f (0) = ± 1
Hence f (x) =
x
2
± 1 ]