MATHEMATICS

/2 2

2 2
= ( t e t )12 – e t dt = 2e4 – e – 
Q.1 If I10 =  x10 sin x dx , then I10 + 90 I8 is 1

0
(A) 10 (/2)6 (B) 10 (/2)9
(C) 10 (/2)8
(D) 10 (/2)7 Q.6 If f(0) = 1, f(2) = 3 and f (2) = 5 then
1
Sol.[B] I10 = (– x10 cos x ) 0 / 2 + 10
/ 2  xf ' ' (2x ) dx is

9
x cos x dx 0
0
/2
(A) 4 (B) 6 (C) 8 (D) 2 [D]
= 10 0 x 9 cos x dx /2

 9
= 10 ( x sin x ) 0  9 0
/ 2
/ 2 
x 8 sin x dx 
Q.7 If I1 =  cos (sin x )dx ,
0
I2 =
 
/2 /2
= 10[(/2)9 – 9I8] I10 + 90I8 = 10(/2)9

0
sin (cos x )dx and I3 =  cos x dx
0
/3
then -
Q.2 If 0 < x < /2, then  cosec x d(sin x) is (A) I1 > I3 > I2 (B) I3 > I1 > I2
/4
(C) I1 > I2 > I3 (D) I3 > I2 > I1 [A]
-
5
1 3
(A) n 2 (B)
2
n
2 Q.8  ([x ]  [ x ])dx
2
=
 sin 3 / 2 
(C) n   (D) None of these
 [B] (where [x] is greatest integer function)
 sin 1 / 2  (A) 3 (B) – 3 (C) 2 (D) – 2 [B]
1 1
e x dx x 2 dx Q.9 The value of
Q.3 Let I1 =  1 x
and I2 =  ex 3
(2  x 3 )
,
0 0
then I1/I2 is-
e 1 3 9  / 4 5 
(A) 3e (B) (C) (D) [A]
3 3e e
 0


 (| sin x |  | cos x | )dx  { x} dx 
1



1
 1– x 2 
Q.4 If   1  x 2  x 4  dx = a nb, then (where [.] denotes greatest integer function and
0
{.} represents fractional part function) is-
1
(A) a = 1, b = 9 (B) a = ,b=3
2 (A) 3 (B) –4 (C) 2 (D) 4 [C]
1 /2 1
(C) a = 2, b = 3 (D) a = ,b=9 [D] x tan 1 x
4 Q.10 If I1 = 0
sin x
dx and I2 = 
0
x
dx
2

e
x2
Q.5 If the value of the integral dx = , then ,
1
e4
I1
the value of 
e
n x dx is then
I2
=

(A) e4 – e –  (B) 2e4 – e – 

(A) 1 (B) 1/2 (C) 2 (D) /2 [C]
(C) 2(e4 – e) –  (D) None
1
Sol.[B] Let lnx = t2 tan 1 x
2 2
x = e t dx = e t . 2t dt
Sol. I2 =  0
x
dx , x = tan

2 2
 
2 2
2 t 2 .e t . dt = t ( 2 te t ) dt
1 1

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 /4 /2 b c
2 1 x
 I2 = 0
sin 2
d =
2 0
sin x
dx = Sol. Note that  f ( x  c) dx
0
=

1 c
. I1
2   f ( x ) dx
b
I1
 =2
I2 1 1
e x dx x 2 dx
/ 4
 x 
2
Q.13 Let I1 =  x 1
and I2 = e x3
(2  x 3 )
,
Q.11 
0
  dx =
 x sin x  cos x 
0 0

3  4 I1
(A) (B)
4 3  then is equal to -
I2
4 4
(C) (D) 3 e 1
4 4 (A) (B) (C) 3e (D)
e 3 3e
[C]
[C]
d  1 
Sol. Since   = 1
dx  x sin x  cos x  x 2 dx
Sol. I2 = e
0
x3
(2  x 3 )
 x cos x
dt
( x sin x  cos x ) 2 {let 1 – x3 = t  x2dx = }
3
integration by parts, 0 1
1 dt 1 et
  x sec x 
/4
4
= 
3 e
1
1 t
(1  t )
=
3e 
0
1 t
dt
I=   tan x  =
 x sin x  cos x 0 4
1 I1
I2 = I1  = 3e
3e I2
Q.12 Which of the following is not correct ? /2
sin 2 x dx
a Q.14 The value of  1 ex
is-
(A) 
a
x . (f (cos x )) 2 .dx = 0  / 2

 
n 
(A) (B)
8 4
 f (cos 
2
(B) x ) dx = n f (cos 2 x ) dx
0 0  
(C) (D) [B]
6 3
nN
/ 2 /2
sin 2 x sin 2 (  x )
 
b c c
Sol. I= dx = dx
(C)  f (x  c) dx =  f ( x ) dx
0 b
 / 2
1 e x
 / 2
1  ex

/ 2
a ex 1
(D)  x.f (sin x ) dx =
I+I= 
 / 2
sin 2 x
1 ex
dx
a
a / 2
 1
  sin
2
f (sin x ) dx [C] I= x dx
2 2
a  / 2

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 / 2  f(x) = x2 + x + 1
or I = 
0
sin 2 x dx f '(x) = 2x + 1
1

2n 
1
2
  (f ( x ) + f '(x) + )dx = 20
1
Q.15 The value of x 
 (sin x )  dx is (where
2
– 2n
10  1
x x  
{x} denotes the fractional part of x) -
  2
Q.17 xe dx is equal to ([x] denotes the
–2n  1 n –10
(A) (B)
2  integral part of x) -
( n  1) 2 n – 1 (A) 0 (B) 1
(C) (D) 2
  (C) 2 (D) 3
1 10  1
2n  x x  

2  2
x  Sol.[A] I = x.e dx =
Sol.[A] I =
 (sin x )  dx
2 –10
– 2n

2 1/ 2
x x
= 2n  (sin x ) dx +
2  (sin x )
2
dx
10  1
0 0 – x   – x   

 (– x ).e  2
dx
–10

–2n  1
=  1   1 
2 Now  – x   =  –  x    1 = –
 2    2 

1 0 0 x  x 
 0 x 2   2x   x 2 
Q.16 Let 6 2    =    1
x  2 

5 4 3
  1  5x  x 2  3  

for at least 3 values of x and f(x) is a 1

If x  odd multiple of .
differentiable function satisfying f(xy) = f(x) + 2
x2(y2–1)+y– 1,  x, y  R and f(1) = 3, then
10  1
 x x 

1 2
I= – x.e  dx = – I  I = 0.
 (f ( x )  f ' (x )   ) dx is equal to - –10
–1

(A) 10 (B) 20 (C) 25 (D) 35 16

 tan
1
Q.18 The value of x  1 dx is
1 0 0 x  x  1
 0 x 2   2 
Sol.[B] 6 2    =  2 x   x  16 4
 (A) + 2 3 (B) – 2 3
5 4 3
  1  5x  x 2  3 3 3
 x   x  4 16
(C) + 2 3 (D) – 2 3
 6 x  2 x 2   2x   x 2  3 3
=   =   Sol.[D] Integrating by parts, the given integral is equal to
5x  4 x 2  3 5x  x 2  3
16
x 1

16
 f(y) = f(1) + y2 – 1 + y – 1 = y2 + y + 1 x tan –1 x 1 
1
x 4 x x 1
{f(1) = 3} 1

dx
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 16 3 9
16 1 dx  x  2x 2 x
=
3
 –
4 
1 x 1
2 2
2
0

(x) = 0 3 0
3 3
16 1 4 t (1  t 2 ) 3 2 6
=
3
 –
4 0
t
dt ( x =1+

t2) 3 9
 x  2x 2 x 0
=
16
3
 –  3 3 =
16
3

–2 3
2
11
2
(x) = 0  0
3
3 2 6
Q.19 The length of perpendicular from the foci S and

x2 y2  11  3
S' on any tangent to ellipse + =1 are 2 
4 9 (x) = 6     x  2x  
'a' and ‘c’ respectively then the value of
 3 2 
1 1
ac   66  3 x 2 2x 3  
 {2x}dx is equal to  (x ) =   x
 3  2 2
 
3  
0
0
 ac

(A) 4 (B) 18 (C) 8 (D) 9  66  3 1 2 

Sol.[C] We know that ac = (semi minor axis)2 = 4 =    
3 2 2 3
Now
9  3  4 176
4 1/ 2 = – 22   =
1  6  3
 {2x}dx  16  {2x}dx  32  4  8
4 0 Q.21 Let f,g and h be continuous function on
[0,a] such that f(x) = f(a–x), g(x) = – g(a–x)
and
1  x  2x 2 x3 1 a

Q.20 If (x) = x  2x 2 x 3 then

3h(x) – 4h(a – x) = 5. Then 0 f ( x ) g(x)

3x  6 x 2 3x  11 9 h(x) dx =
(A) 5/4 (B) 3/4 (C) 1 (D) 0
1
0 ( x ) dx is Sol.[D]
a
 0 f ( x ) g(x) h(x) dx
176 176 a
(A)
5
(B) –
3
=  0 f (a  x ) g(a – x) h(a–x) dx

186 a  3h ( x )  5 
(C)
3
(D) None of these =– 0 f ( x ) g(x) 
 4  dx

Sol.[B] R2  R2 – R1 3 a 5 a
R3 = R3 – 3R1 = –
4 0 
f ( x ) g(x) h(x) dx +
4 0
f (x) 
1  x  2x 2 x 3 1 g(x) dx
(x) = 1 3 2
7 a 5 a
3 2 6
4 0 
f ( x ) g(x) h(x) dx =
4 0
f ( x ) g(x) 
dx = 0
R2 {f(a – x) g(a – x) = – f(x) g(x)}
R1 = R1 –
2 a
So 0 f (x) g(x) h(x) dx = 0
2
1  x  2x  1/ 2 x  3  3/ 2 0
(x) = 1 3 2
3 2 6

R3
R2 = R2 –
3
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Q.22 Let f be a real valued function satisfying x
 x4 x2 
p'(x) =  0
p''(x) dx + 3 = a 
 4
 +
2 
x 12

f(x) + f(x + 6) = f(x + 3) + f(x + 9). Then 

x
f(t) 3

Since p(1) = 1, we get

dt is
x
(A) a linear function of x p(x) =  p'(x) dx + 1
(B) an exponential function of x 1

(C) a constant function  x5 x3 7 

= a    + 3 (x – 1) + 1
(D) None of these  20 6 60 
Sol.[C] f(x) + f(x + 6) = f(x + 3) + f(x+ 9) …(1)
60( 3  1)
x=x+3 p(– 1) = – 1, so a =
7
f(x + 3) + f(x + 9) = f(x + 6) + f(x + 12) x x
3 1
n
from eq (1) 1
p(x) dx = 
1
7
(3x5 – 10x3) + x
f(x) + f(x + 6) – f(x + 9) + f(x + 9)
= f(x + 6) + f(x + 12) 3 ]dx

f(x) = f(x + 12) 3 1  x6 5 4  2

  x  + x
1
=  2  3
x 12 7  2  2
0

Let g(x) =  f(t) dt

3 1  1  5  2
x
  + 3 3 3
= = +
7 2 2 2 14 7
g'(x) =  f ( t ) xx 12

g ' (x) = f(x + 12) – f(x) = 0

/2
So g(x) is a constant function. Q.24 If  0
log sin x dx = k, then


Q.23 Let P(x) be a polynomial of least degree whose
graph has three points of inflection (–1, – 1),  log (1 + cos x) dx is given by
(1, 1) and a point with abscissa 0 at which the 0
curve is inclined to the axis of abscissa at an
1
(A)  log 2 + 4k (B)  log 2 + 2k
angle of 60º. Then 
0
P(x) dx equals to - (C)  log 2 + k (D)  log 9 + k2


(A)
3 34
14
(B)
3 3
7
Sol.[A] I = 
0
log (1 + cos x) dx

3 7 32 
(C) (D) x
14 7
Sol.[A] Required function is a polynomial, the abscissa
=  0
log (2 cos2
2
) dx

of the points of inflection can only be among  

x
the roots of the second derivative = 
0
log 2 dx + 
0
log cos2
2
dx

p''(x) = ax (x – 1) (x + 1) = a(x3 – x)

x
at the point x = 0, p'(0) = tan 60º = 3 =  log 2 + 2 
0
log cos
2
dx

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 DEFINITE INTEGRATION 5
 /2 Since sin x < x < tan x
=  log 2 + 4 
0
log cos xdx

sin(sin x )
>
sin x
>
sin(tan x )
for
sin x x tan x
/2

=  log 2 + 4 0
log cos (/2 – x) dx 
6
<x<

3
/2  I2 > I1 > I3
=  log 2 + 4 0
log sin x dx

dx
=  log 2 + 4k Q.27 If I =  x2 1 x
, then I equals
1

(A) 2 + log ( 2 –1)

1
2n 
2 x  (B) 2 –log ( 2 –1)
(sin  x)   dx is (where
Q.25 The value of
 2n
 2
(C) log ( 2 + 1) + 2

{x} denotes the fractional part of x) - (D) 2 + log (3 – 2 )

2 n   1 n
(A) (B)
 2  Sol.[C] Put 1 + x = t2, then
( n  1) 
1 2t
(C)

(D) I=  dt
( t 2  1)
t
2
Integrating by parts, we get
2 n  1  
1  dt
2 I=–
(t 2 
 1) t 

–  (t 2
 1) t 2
2 2
1 
2n 
x  1  1 1
2
= –   2  2  dt
Sol.[A] I =

2n
(sin x)   dx
2 2 2
 t 1 t 

1 1 t 1 1 
= – 
 2 log t  1  t 

2 1/ 2 2  
x x 2
= 2n  (sin x)
2
dx +  (sin x)
2
dx 1 1 1 1/ t 
0 0 = –
log  +0
2 2 1  1 / t  t 
2n  1 1 2 1 1
= + log +
2 2 2 1 2
/3 /3 = 2 – log ( 2 –1)
sin x sin(sin x )
Q.26 Let I1 = 
/6
x
dx, I2 =
/6
 sin x

x dx
dx, Q.28 If I =  (1  x ) (2  x ) (3  x ) , then I
0
/3
sin(tan x )
I3 =  tan x
dx, then - equals-
/6

(A) (2 2 – 3 –1)
(A) I1 < I2 < I3 (B) I2 < I1< I3 2
(C) I3 < I1 < I2 (D) I3 < I2 < I1 
(B) (2 2 + 3 –1)
sin x 2
Sol.[C] f(x) = is a decreasing function and
x 
(C) (2 2 – 3 +1)
sin x 2
> 0 for all x in (0,)
x
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 (D) None of these 31 35
(A) (B)
Sol.[A] Put 2
x = t or x = t , so that 2 2

t2
37 39
(C) (D)
I= 2 dt 2 2
0
(1  t 2 ) ( 2  t 2 ) (3  t 2 )
Sol.[A] We can write

 1 4 3 
=  2
 2
  dt I = I1 + I2 + I3
0
1 t 2t 3 t2 
2
= where I1 =  | x  1 | dx etc.
3

 Put x + 1 = t, so that
     
  tan 1 t  4 tan 1  t   3 tan 1  t  
     3 0 3
 2  2 3  3   0
I1 =  | t | dt =  ( t ) dt + t dt
2 2 0
  
+2 2  – 3  
0 3
=– 1 2 1  13
2 2 2 =– t  + t2 =
2  2 2 0 2
 9
= (2 2 – 3 –1) Similarly, I2 = I3 =
2 2
1
1 x 31
Q.29 If I = x 1 x
dx, the I equals- Thus, I =
2
0

  Q.31 If I =
(A) 1 + (B) 1 – (C)  (D) 
4 4
0

 [x
3
– 2  3x 2  3x  ( x  1) cos( x  1)] dx
2
Sol.[B] We can write
1
, then I equals:
x (1  x )
I=  1 x 2
dx (A) –4 (B) –3
0
(C) –2 (D) –1
1 
2
x 1 1 x Sol.[C] We can write
=   1 x2

1 x2
  dx
2 
1 x 
I=
0 0

 [( x  1)
3
 
1  1  ( x  1) cos ( x  1)] dx
=  1  x 2  sin 1 x 

  0 2
1
x 1  Put x + 1= t, so that
+  1 x2  sin 1 x 
2 2  0 1

 [t
3
I=  1  t cos t ] dt
 1   1
=– +1+  =1–
2 2 2 4 1
=  (1) dt = – t ]11 = –2
1
2
Q.30 If I =  (| x  1 |  | x  2 | | x  1 |) dx , then as t3 + t cos t is an odd function
3

I equals:

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Q.32 If I = 1 1
 =1– – +e–e+1
|cos x |  1  1  e e
 e 2 sin  2 cos x   3 cos  2 cos x 
0
= 2(1 – 1/e)
sin dx, then I equals
sin 2kx
Q.34 If for k  N, = 2[cos x + cos 3x +......
(A) 7 e cos (1/2) sin x
(B) 7 e [cos (1/2) – sin (1/2)] + cos (2k –1) x],
(C) 0
(D) None of these / 2

Sol.[D] Put
1
cos x = t, so that –sin x dx = 2dt and
then value of I =  sin 2kx cot x dx is
0
2
1 / 2
(A) –/2 (B) 0
(C) /2 (D) 
e
|t|
I= (2 sin t + 3 cost) (–2) dt / 2
sin 2kx

1/ 2
Sol.[C] Writing I = cos x dx
0
sin x
As e|t| sin t is an odd function, and e|t| cos t is an
even function, and using the given identity, we can write
/ 2
1/ 2
I=6 e
t
cos t dt = 6et cos t ]10/ 2 + 6 I=  2[cos x + cos 3x + ......+ cos (2k –1) x]
0
0
1/ 2 cos x dx
e
t
sin t dt
/ 2


0
= [(1 + cos 2x) + (cos 4x + cos 2x) + ..... +
 1 
I = 6  e cos   1 + 6et sin t ]10/ 2 – 6 0

 2  (cos 2kx + cos (2k –2) x)] dx

1/ 2
=
e
t
cos t dt
/ 2
0  1 1 
 x  sin 2 x  sin 4 x  .....  (sin 2kx ) 
  2 2k  0
1 1 
 7I = 6 e  cos    sin    1
 2 2  
=
2

b
e
dx xa
Q.33 If I =  | log x | , then I equals Q.35 If b > a, and I =  bx
dx, then I equals
1/ e x2 a

(A) 2 (B) 2/e 

(A) (b – a) (B)  (b – a)
(C) 2( 1 – 1/e) (D) 0 2
(C) /2 (D) 2 (b – a)
1
Sol.[C] Put = t, so that
x Sol.[A] Put b –x = t2, so that
0
1/ e
1 b  t2  a
I=  log (–1) dt
t
I=  t2
(–2t) dt
e b a
e
=   log t dt c

1/ e =2  c 2  t 2 dt where c = ba
1 e 0
=  ( log t ) dt +  (log t ) dt 1
= 2  t c2  t 2 
c2  t 
sin 1  
c

1/ e 1
 2 2  c  0

= (–t log t + t )] ]11 / e + (t + log t –t) ]1e

= 0 + c2 sin–1 (1) – 0
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  
= (b –a) (A) (B)
2 sin 

b
|x| 
Q.36  x
dx equals: sin 
a

(A) |a| – |b| (B) |b| – |a| 2  

(C) (D) None of these
sin 
(C) ||a| – |b|| (D) |b – a|
1 1
Sol.[B] Case -1 dx  sin 1 ( x sin  ) 
b b
Sol.[B]  1  ( x sin  ) 2
= 
 sin 

 0
0
|x|
If a < b < 0,  x dx = –  dx
a a sin 1 (sin  )  0 
= =
= – (b –a) = |b| – |a| sin  sin 

(a < b < 0; |b| = – b and |a| = –a)  3

(for
 << sin–1 sin  = – )
Case- 2 2 2
b 0 b  
|x| = (For given interval =
If a < 0 < b,  x dx = –  dx +  dx sin  sin 
a a 0
= a + b = |b| – |a| (a
 < 0, |a| = –a)

)
Case- 3 sin 
b
|x| 50
If 0 < a < b,  x
dx Q.39  [arg | z  1 |] dx , where | | is modulus and
a
0
b
=  dx = b – a = |b| –|a| [ ] is greatest integer function is-
a (A) 0 (B) 25 (C) 50 (D) 100
Sol.[A] |z –1| is a positive real number
/2
1  cos 2 x
Q.37  2
dx equals: arg |z –1| = 0
 / 2 50 50

(A) 1 (B) 2 (C) 1/2 (D) 0   [arg | z – 1 |] dx =  0 dx  0

0 0
/ 2 /2
1
Sol.[B] I =  | sin x | dx = 2  | sin x | dx (Even Q.40  max {x, x
3
} dx equals-
 / 2 0
1
function) (A) 2 (B) 1/4 (C) 3/4 (D) 3/2
/2 /2 1 0 1
=2  sin x  –2 [cos x ]0 =2 Sol.[B]  max{x, x } dx =
3
 x dx +
3
 x dx
0
1 1 0
1
dx 1 1 1
Q.38 If /2 <  < 3/2, then  2
1  x sin  2
is =
4
+
2
=
4
0
2
equal to - Q.41  min {| x |, [ x ]} dx equals:
2
(A) –2 (B) 0 (C) 1 (D) –1
Sol.[A] Clearly min. {|x|, [x]} = [x]

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 2 1 0 16 16
e sin y
  [ x ] dx =  [ x ] dx +  [ x ] dx + I=  y
dy = |F(y)|  = F(16) –F(1)
2 2 1 1 1

thus one of the possible value of k = 16

1

 [ x ] dx + Q.44 If I = 
2 x (1  sin x )
dx, then I equals-
0

1  cos 2 x
2
(A)  (B)
 [ x ] dx 2 2
1  
(C) (D) –
1 0 2 2 2
=–2  dx –  dx + 0 +  dx
Sol.[B] We can write
2 1 1
I = I1 + I2
= –2 –1 + 1 = –2

x 2x
Q.42 If g(x) =  cos t dt , then g(x + ) equals-
4 where I1 =  1  cos 2 x dx =

0
0
(A) g(x) + g() (B) g(x) – g()

(C) g(x) . g() (D) 0 x sin x
and I2 = 4 
x 0
1  cos 2 x
 cos
4
Sol.[A] g(x + ) = ( t ) dt
=
0
x x 

 cos 4 ( t ) dt +  cos
4 
= t dt (  x ) sin(  x )
0 x 4 dx
1  (cos(  x )) 2
... (3) 0
x 
=
 cos 4 t dt +  cos
4
= t dt (cos4 t is
0 0
periodic function) 
dx
4   I2
1  cos 2 x
= g(x) + g() 0


cos ec 2 x
Q.43 Let
d
F(x) =
e sin x
, x > 0. If
 I2= 2  cos ec 2 x  cot 2 x dx
0
dx x

cos ec 2 x dx
4
2e sin x
2
= 2  1  2 cot 2 x

0
dx = f(k) – f(1) then one of the
x
1 =
possible value of k is:

(A) 2 (B) 4 (C) 8 (D) 16  2
2
tan 1  
2 cot x 
0
4 2 4 2
2e sin x 2 xe sin x
Sol.[D] Let I =  x
dx =  x2 = –
  
2    =
1 1
 2 2
Put x2 = y, then 2x dx = dy
2
x 1 4
y 1 16

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 log e 5 ex ex – 1
Questions Add (24–6-09) Q.4  0 ex  3
dx =

(A) 3 + 2 (B) 4 – 
(C) 2 +  (D) 4 +  

x 2 dx
Q.4  (x 2
 a 2 ) (x 2  b 2 ) (x 2  c 2 )
= Sol.[B] Put ex – 1 = t2
0
ex dx = 2tdt

2(a  b)(b  c)(c  a ) 2 t .2 tdt
I =  0 t2  4

dx
then  (x
0
2
 4)( x 2  9)
=
n 1
Q.5  n
f ( x )dx = n2 + 1 n Z then
  
(A) (B) (C) (D) 3
60 20 40  –2
f ( x ) dx =
 (A) 10 (B) 13
80 (C) 15 (D) 18 
Sol.[A] put c = 0, a = 2, b = 3 Sol.[C] I =
–1 0 1
 –2
f ( x ) dx   –1
f ( x ) dx  0
f ( x ) dx 
/ 2

x
n
Q.5 If In = sinx dx then I7 + 42 I5 = 2 3

0  1
f ( x )dx  2
f ( x )dx
7 6
 
(A)   (B)   1000
2 2
 
n
Q.6 e x –[x] dx =
6 7 n –1
  n 1
(C) 7   (D) 7  
2 2 e1000 – 1 e1000 – 1
/2 (A) (B)

Sol.[C] In= x
n
sinx dx =
x n
( cos x )  / 2
0 –
1000 e –1
e –1
0
/2 (C) 1000 (e – 1) (D) 
1000
 nx
n 1
n 1
(– cos x)dx
0
/2
Sol.[C]  n –1
e{ x}dx = n – (n – 1)  0
e{x}dx


1

=n xn–1 cosx dx =  0 e x dx = e –1
0
= n
/2 Q.6 If C0, C1, C2,....., Cn are their usual meaning,
[[x n 1
(sin x )]0 / 2  (n  1) x
n 2
sin x dx ] C0 C1
0 then – +
n 1
n (n  1) (n  1) (n  2)

In = n   – (n – 1)n In–2 put n = 7 C2
2 + .... to (n+1) terms is equal
(n  2) ( n  3)
4 to–
Q.6  ( | x  1 | 3 | x  2 |  4 | x  3 | ) dx  1

x
n 1
0 (A) (1  x ) n 1 dx (B)
(A) 37 (B) 14 0
(C) 22 (D) 11 1

x (1  x ) n 1 dx
n
Sol.[A] Break integral at x = 1, 2, 3 0

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 1 sin  z
(C)  x (1  x ) dx
n 1 n 1
(D) None B= 1 ( z  1)2 dz
0
of these sin  z
Sol.[C] (1–x)n = C0 – C1x + C2x2 – C3x3 + .......+ (–1)n B=– 1 2
( z  1)
dz zt
Cnxn
Multiplying by xn–1 to both side sin  t
xn–1 (1 –x)n = C0 xn–1 –C1xn + C2xn+1 .....
...+ (–1)n Cnx2n–1
B=– 1 2
t 1
dt = – A

Now multiplying with (1 – x) to both sides B=–A A+B=0

n–1 n+1 n–1 n n +1
x (1 – x) = (C0 x – C1 x + C2x ....... A A2 A
+ (–1) Cnx n 2n–1
) (1 – x) Now  = 1 A2 1 =0
1 2A 2 1
Now integrating w.r.t. x from 0 to 1
1
C0
x
n 1
(1  x ) n 1 dx = –
n (n  1) x tan 1 ( t )
0 Q.1 If f(x) =
 1 t
dt xR+ then the

 1 
C1 value of f(e2) – f  2  is
(n  1) (n  2)
+ e 

(A) 0 (B) (C)  (D) 2
C2 2
.........
(n  2) ( n  3) x tan 1 ( t )
sin 
Sol.[C] f(x) 
1 t
dt
t dt
Q.6 If A = 1 1 t2
and B =
f
1
 

1/ x tan 1 ( t )
dt
x 1 t
cos ec dt put t = 1/u
1 t (1  t 2 )
then the value of dt = –1/u2 du
1
tan 1    1 
f(1/x)  = x  u    2  du
A B
A A2 B  1 1/ u
 u 
e e B2  1 is
1 A  B2
2
1 1
tan 1  
f(1/x)  – x u = –
(A) sin
sin 
(B) cosec (C) 0 (D) 1

1 u
t dt
Sol.[C] A = 1 1 t2
x cot 1 ( u )
cos ec dt 1 1 du
B = 1 t (1  t ) 2 Put t =
z
, dt = – u

x cot 1 ( t )
1
=–
1 t
dt

dz
z2 x tan 1 t  cot 1 t
1
f(x) – f(1/x) =
 1 t
dt

 dz x  1
B= 1
sin 
z2 =
1 2  dt
t
1 1  
1  2  f(x) – f(1/x) = log(x)
z z  2

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  2 2
logee2 = 
 e
2
f(e2) – f(1/e2) = t2
2t 2 e t
2
= dt = ( te t )12 – dt
2
1 1

= 2e4 – e – a
1
Q.4  1
[ x[1  sin x ]  1] dx is ([·] denotes
y x2
sin t dy
greatest integer function)
 cos t 
2
Q.8 If dt = dt. Then is
(A) 2 (B) 0 (C) 1 (D) None 0
t dx
0
Sol.[A] –1 < x < 0 0 < 1 + sin  x < 1
equal to -
[x [1 + sin  x] + 1] = 1
2 sin x
[1 + sin  x] = 0 (A) (B)
x cos y
0<x<1 1 < 1 + sin x < 2
[1 + sin  x] = 1 2 sin x 2
x cos y
[x [1 + sin  x] + 1] = 1
1 1 2 sin x 2 2 sin x 2
1dx 2 0 dx = 2[ x ]10 = 2 (C)
x cos y 2
(D)
cos y 2

Sol.[C] On differentiating under the integral sign,

Q.4 Let [x] stands for the greatest integer function. dy sin x 2
21 we get (cos y2) = .2x
dx x2
 [x]
3
The value of dx is -
0 dy 2 sin x 2
(A) 44001 (B) 44000  =
dx x cos y 2
(C) 40400 (D) None of these
21 20 r 1
Sol.[D]  [ x ]3 dx =   [x]
r 0
3
dx Q.6 Number of solution for x between 3 and 15 if
x [x]
0 r
20 r 1  [ t ]dt   t dt , where [.] denotes
=  r
r 0
3
dx 0 0
greatest integer function, is -
r
20 2 (A) 12 (B) 13
 20.21 
= r
1
3

 2 
 = 44100 (C) 11
x
(D) infinitely many
[x] [ x ]{ x }

Sol.[A] 
0
[ t ]dt = 
0
[ t ]dt +  [ t ]dt
[x]
=
2

e
x2
Q.2 If dx = a, then the value of
1 [x]

 t dt
0
e4


e
n ( x ) dx is-
[ x ]{ x } [x] [x]

(A) e – e 4
(B) e – a 4  
[x]
[ t ] dt = 
0
t dt –  [ t ] dt
0
(C) 2e4 – a (D) 2e4 – e – a
e4 [x]
Sol.[D] Let I = e
n ( x ) dx. =  {t} dt
0

Put n x = t  dx = 2t e
2 t2 dt

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 1 1 1
{x} [x] = [x] .
2
i.e. {x} =
2 Q.6  | 3x  1 | dx equals-
0
1 1 1 (A) 5/6 (B) 5/3
thus 3 < x = n + < 15 i.e. 3 – < n < 15 –
2 2 2 (C) 10/3 (D) 5 [A]
 n can take 12 values.
2
No. of solutions is 12. x
Q.7  3 x  x
dx is equal to-
1
/ 4 (A) 2/1 (B) 3/4
 tan
2
Q.1 x dx equals- (C) 1/2 (D) None of these [C]
0
(A) /4 (B) 1 + (/4)

(C) 1– (/4) (D)1– ( /2) [C] x sin x
Q.8  1  cos 2 x dx equals-
0
e
1 (A) 0 (B) /4
Q.2 The value of  x
dx - (C) 2/4 (D) 2/2 [C]
1
(A)  (B) 0
/ 2
(C) 1 (D) log (1+e) [C] dx
Q.9  1  cot x
equals-
0

 1 
e tan x (A) 1 (B)
Q.3  1 x2 dx equals- 4
0 1 
(C) (D) [B]
(A) 1 (B) e/2 +1 2 2
(C) e/2 – 1 (D) None of these [C]
1
dx
/ 4
sec x 2
Q.10  ( x 2  2 x  2) 3 =
0
Q.4  (1  tan x ) ( 2  tan x )
dx equals-
3  8  1
0
(A) (B)
2 32 4
(A) loge (B) loge 3
3 (C) 0 (D) None of these [A]

1 4 4
(C) loge (D) loge 3
2 3 3 x 2 sin x
[D]
Q.11  1 x6
dx equals-
3
(A) 4 (B) 2
(C) 0 (D) None of these [C]
 2
 x , when 0  x  1
Q.5 If f(x) =  , then

 x , when 1  x  2
1
2

 f ( x ) dx equals-
Q.12 ( 1  x  x 2  1  x  x 2 ) dx equals-
1
0
(A) 1 (B) 0
1 1
(A) (4 2 – 1) (B) ( 4 2 + 1) (C) 2 (D) 2 [B]
3 3
(C) 0 (D) does not exist [A]
2

 cos
4
Q.13 x dx equals-
0

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 (A) 3/8 (B) 3/4 1 1 1 1 
Q.20 lim     .....  
(C) 3/2 (D) 3 [B] n 
 n n 1 n  2 3n 
equals-
2
sin 2 (A) log 2 (B) log 4
Q.14  a  b cos  d equals- (C) 0 (D) loge3 [D]
0
(A) 1 (B) 2
(C) /4 (D) 0 [D] Q.21 lim
n 
 12 22 r2 1 
400   3 3
 3 3
 .....  3 3
 .....  
1  n 2 n r n 2n 
Q.15  1  cos 2 x dx is equal to-
=
0
(A) (1/2) log 3 (B) (1/3) log 2
(A) 400 2
(C) 3 log 2 (D) (1/2) log 2 [B]
(B) 800 2
(C) 0 m m m m
Q.22 lim 1  2  3  .....  n = ……; [m > –
(D) None of these [B] n 
n m 1
1]
1
1000 (A) (B) 0
x [ x ] m 1
Q.16 The value of e dx is - (C) (m +1) (D) None of these [A]
0

e1000  1 e1000  1 1 1 1
(A) (B) Q.23 The limit of the sum + + +
1000 e 1 n n 1 n  2
e 1 1 1
(C) 1000(e – 1) (D) [C] +……..+ , when n is indefinitely
1000 n 3 3n
increased
(A) n 3 (B) 0
/2
(C) n 2 (D) None of these [A]
Q.17  log cos x dx equals-
0 2
dx
(A) (/2) log (1/2) (B) log 2 Q.24 If a <  10  3 cos x < b then the ordered
(C) –log 2 (D) 2log 2 [A] 0

pair
/ 2 (a, b) =
 sin
7
Q.18 x cos x dx equals-  2 2   
0 (A)  ,  (B)  , 
 13 7  3 2
(A) 1/7 (B) 1/8
(C) /16 (D) /14 [B] (C) (0, ) (D) None of these [A]
If a
/ 2 1
x2
Q.19  sin x dx equals-
5 Q.25 If e ( x   ) dx  0 then-
0 0

(A) 8/15 (B) 4/15 (A) 1 <  < 2 (B)  < 0

8  8 (C) 0 <  < 1 (D)  = 0 [C]
(C) (D) [A]
15 15
2 2
dx dx
Q.26 LetI1 =  2
and I2 =  x
then-
1 1 x 1

(A) I1 > I2 (B) I2 > I1

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 (C) I1 = I2 (D) I1 > 2I2 [B] 3
Q.32 The value of  (| x  2 | [ x ]) dx is ([x]
1
Q.27 Let f(x), g(x) & h(x) be continuous function on
stands for greatest integer less than or equal to
[0, a] such that f(x) = f(a – x), g(x)= – g(a – x),
x)
a
(A) 7 (B) 5
3h(x) – 4h(a – x) = 5. then  f(x) g(x) h(x) dx
(C) 4 (D) 3 [A]
0
=
(A) 1 (B) 0 Q.33 Let f : R  R, g : R  R be continuous
(C) a (D) None of these [B] functions. Then the value of integral
 x2 
 
y
1 d2y n1 / 
f  4  [f ( x )  f (  x )]
 
Q.28 If x =  dt then =
 dx is
0 1  4t 2 dx 2 n  x 
2
 
g  [g ( x )  g (  x )]
(A) 2y (B) 4y  4 
(C) 8y (D) 6y [B] (A) dependent on  (B) a non- zero constant
(C) zero (D) None of these [C]
Q.29 If f(x) is a continuous function and f(0) = 0, then
  d
x  Q.34 If f(x) = g(x) for a  x  b then,
lim 
 f ( u ) du 
 is - dx
x 0  0 
x b
 
   f ( x ) g ( x ) dx equals
(A) 1 (B) 0 a

(C) 2 (D) None of these [B] (A) f(b) – f(a)

B) g(b) – g(a)
b
sin ( x  a )  cos ( x  a ) [f (b)]2  [f (a )]2
Q.30  sin (b  x )  cos (b  x ) dx is equal to (C)
2
a
- [g ( b)]  [g (a )]2
2
(D)
b
sin ( x  a )  cos ( x  a ) 2
(A)  sin (b  x )  cos (b  x )
dx [C]
a
b Q.35 If f(0) = 1, f(2) = 3, f (2) = 5 and f  (0) is finite,
sin ( b  x )  cos (b  x )
(B)  sin ( x  a )  cos ( x  a ) dx 1

a then  x.f ( 2x )dx is equal to

0
b
sin ( b  x )  cos ( b  x ) (A) zero (B) 1
(C)  dx
a
sin ( x  a )  cos ( x  a ) (C) 2 (D) None of these [C]
(D) None of these [B] 3

 f (cos
2
Q.36 Let I1 = x ) dx , I2 =
1 0
Q.31 If In =  xn e–x dx for n  N, then I7 – 7I6 = 2 

 f (cos x ) dx and I3 =  f (cos

2 2
0
x ) dx
0 0
1 1 , then
(A) – (B) –
2e e (A) I1 + 2I3 = 3I2 (B) I1 = 2I2 + I3
1 (C) I2 + I3 = I1 (D) I1 = 2I3 [C]
(C) (D) e [B]
e
Q.37 The value of the function

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 x 1 5 7 
 (n an 2  2 a  2 b 
2
f(x) =1 + x + t  2nt ) dt where f (D) 2 2
1
a b  
[A]
(x) vanishes is
(A) e–1 (B) 0
(C) 2e–1 (D) 1 + 2e–1 [D] 3
Q.5 If y = (x)[x] where x = [x] + (x), then  ydx =
3n 0
n
Q.38 lim
n   2
 n2
is equal to (A) 2/3 (B) 5/6 (C) 1 (D) 11/6 [D]
r  2 n 1 r

2 3
(A) log (B) log Q.6 If the tangent to the graph of the function
3 2 y = f (x) makes angles of /4 and /3 with the
2 3 x-axis at the points x = 2 and x = 4 respectively,
(C) log (D) log [B]
3 2 4
then the value of  f '(x) f "(x) dx =
1 1 2
2 2
2 x
Q.1 If  + 2  x e dx =  e  x dx , then the (A) 0 (B) 2 (C) 3 (D) 1 [D]
0 0
value of  is -  
x 2   ax 2
(A) e (B) 1 Q.7 If e .dx = , then e dx , a >
(C) 0 (D) 1/e [D] 0 2 0
0 is-
n 4  
ex ex  3 (A) (B)
Q.2 The value of  ex  2
dx is- 2 2a
n 3
2  1 
4  (C) (D)
(A) (B) 4 – a 2 a
2 2 [D]
2
(C) 2 –  (D) [A]
2 Q.8 Let a, b, c be non zero real number such that
4 4 1
Q.3 If  f ( x ) dx  4 and  [3 – f(x)] dx = 7,  (1 + cos8x) (ax2 + bx + c)dx
1 2 0
1 2
then the value of  f (x) dx is- = (1+cos8x)(ax2+ bx + c)dx, then the
2 0
(A) 2 (B) – 3 quadratic
(C) – 5 (D) none of these [C] equation ax2 + bx + c = 0 has-
1 1 (A) No root in (0, 2)
Q.4 If for non-zero x, af(x) + bf   = – 5 (B) At least one root in (0, 2)
x x
2 (C) A double root in (0, 2)
where a  b then  f(x) dx = ........ (D) Two imaginary roots [B]
1

1  7  Q.9 The value of

(A) 2 2 an 2  5a  2 b 
a b   2 4/3
n sin ( 2 x  3) 2
 7   dx –3 
(B) (a2 – b2) an 2  3a  b  1 x 7/3
 2  2
 11 
 3 7  n sin   2x 
(C) (a2 + b2) an 2  a  b   3  dx is -
 2 2 
(1  3x )

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 (A) 1 (B) 0 x
log t 1
(C) 2 (D) none of these [B] Q.16 If f(x) =  1 t dt, then f(x) + f   = k(log
x
1

x log x x)2;then k equals to-
Q.10  (1  x 2 ) 2 dx -
0 (A) 1 (B) 1/2
(A) 0 (B) 1 (C) 1/4 (D) none of these [B]
(C) 1/2 (C) none of these [A]
x
Q.11 Let f(x) be a continuous function on [0, a] such Q.17 f(x) =  t(t – 1)(t – 2) dt takes on its minimum
that f (x) f (a – x) = 1 where a > 0, then 0
a value when
dx
 1  f (x) = (A) x = 1, 2 (B) x = 0, 1
0 (C) x = 0, 2 (D) none of these [C]
(A) 0 (B) a
(C) a/2 (D) none of these [D] 1
Q.18 If p =  tan–1x dx and
0
 
Q.12 If In =  e–x.xn–1.dx then  e–x xn–1 dx is 1
0 0 q = cot–1(1 – x + x2) dx then the value of
equal to - 0

1 p
(A)  In (B) .In is
q

(A) 1/4 (B) 1/2 (C) 1/8 (D) 1 [B]
In
(C) (D) n.In [C]
n / 2

( 2 x  3) sin x Q.19  cos 2 n 1 x  cos 2 n 1 x dx, n  N -
Q.13  1  cos 2 x
dx -  / 2
0
2 4
 (A) (B)
(A) ( + 3) (B) ( – 3)  n 1 2n  1
2 (C) 0 (D) none of these [B]
(C) ( + 3)  (D) none of these [A]
1
16  / 3
Q.14 The value of  | sin x | dx is -
Q.20 The value of  [x [1 + sin x] + 1] dx is,
1
0
where [ ] denotes greatest integer function -
(A) 21 (B) 16/3 (C) 32/3 (D) 21/2 [D]
(A) 3 (B) 2
(C) 8 (D) 1 [B]
Q.15 The tangent to the curve y = f(x) at the point
with abscissa x = 1 form an angle of /6 and at
the point x = 2 an angle of /3 and at the point Q.21 Let f(x) and g(x) be two functions satisfying
x = 3 an angle of /4. If f '' (x) is continuous, f(x2) + g(4 – x) = 2x3 ; g(4 – x) + g(x) = 0, then
3 3 4
then the value of  f ''(x) f '(x) dx +  f ''(x) the value of  f(x2) dx =
1 2 4
dx is (A) 128 (B) 64
4 3 1 3 3 1 (C) 256 (D) none of these [C]
(A) (B)
3 3 2
/ 2
43 3 cos x
(C) (D) none of these [C] Q.22 The value of  x
dx is equal to -
3  / 2 1  e
(A) 0 (B) 1
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 (C) 2 (D) none of these [B] / 2


dx
 [f(x) + f(–x)] [g(x) –g(–x)] dx is-
 / 2
Q.23 If  25 cos 2 x  36 sin 2 x = m, then m is -
[IIT-1990]
0

1 1 (A)  (B) 1 (C) –1 (D) 0 [D]

(A) (B)
30 15 / 2
dx
(C)
1
(D) –
1
[A]
Q.4 The value of  1  tan 3 x
is: [IIT-
10 30 0
1993]
(A) 0 (B) 1 (C) /2 (D) /4 [D]
Q.24 lim
n 
2

 1 
2 / n2

1  2 2 

4/ n2

1  3 2 

6/ n2
 n 2 
....1  2 
2n / n 2 

Q.5 The value of  [2 sin x ] dx where [.]
1  2   2   2   0
 n   n   n   n  
  represents the greatest integral function is :
is equal to - [IIT-1995]
(A) e/4 (B) 4/e 5
(C) 1 (D) none of these [B] (A) – (B) –
3
5
(C) (D) – 2 [A]
3
Q.25 Evaluate :
lim 1
n  Q.6 If f(x) = A sin (x/2) + B, f    = 2 and
1/ n 2
  2 3 n 
 tan 2n . tan 2n . tan 2n ........ tan 2n  1
  2A
(A) 0 (B) 1
 f(x) dx =

, then the constants A and B
0

(C) 2 (C) 2/ [B] are

[IIT 1995]
Q.1 For any integer n the integral (A) / 2 and / 2 (B) 2/and 3
 (C) 0 and – 4/ (D) 4/and 0 [D]
cos 2 x
e cos3 (2n +1) x dx has the value
0 Q.7 Let f be a positive function, let
[IIT-1985] k k
I1= 1 k x. f [x (1–x)]dx& I2=  f [x (1
(A)  (B) 1 1 k

(C) 0 (D) None of these [C] – x)]dx,

I1
Q.2 Let f : R  R be a differentiable function and where (2k–1) > 0 , then is - [IIT 1997]
I2
f (x)
2t (A) 2 (B) k
f(1) = 4 then the value of lim
x 1  dt
4
x 1 (C) 1/2 (D) 1 [C]
is- 2n
lim 1 r
[IIT-1990] Q.9 n 
 equals- [IIT 1997]
(A) 8f (1) (B) 4f (1) n r 1 n  r2
2

(C) 2f (1) (D) f (1) [A] (A) 1 + 5 (B) –1 +

5
Q.3 Let f : R  R and g : R  R be continuous
(C) –1 + 2 (D) 1 + 2
functions then the value of the integral
[B]
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 3 1
x 1
(A) – g(2) < (B) 0  g(2) < 2
2 2
Q.10 If  f(t) dt = x +  tf (t) dt, then the value of
3 5
0 x (C) < g(2)  (D) 2 < g(2) < 4 [B]
f(1) is – [IIT 1998] 2 2
(A) 1/2 (B) 0
(C) 1 (D) – 1/2 [A] e2
log e x
Q.16 The value of integral  x
dx is-
Q.11 Let f(x) = x –[x] for every real number x, where e 1

1 [IIT-2000]
[x] is the integral part of x. then  f ( x ) dx (A)
3
(B)
5
(C) 3 (D) 5 [B]
1 2 2
is:
[IIT-1998]

e
cos x
sin x for | x |  2
(A) 1 (B) 2 Q.17 If F(x) =  . Then

 2 otherwise
(C) 0 (D) –1/2 [A]
3
Q.12 If for a real number y, [y] is the greatest integer  F( x ) dx = [IIT- 2000]
less than or equal to y, then the value of the 2
(A) 0 (B) 1 (C) 2 (D) 3 [C]
3 / 2
integral  [2 sin x] dx is - [IIT 1999]
e
x
/ 2 Q.18 Let f(x) = ( x  1)( x  2) dx . Then f
(A) –  (B) 0 decreases in the interval- [IIT 2000 Scr]
(C) – 2 (D) /2 [C] (A) (–, –2) (B) (–2, –1)
(C) (1, 2) (D) (2, +) [C]
3 / 4
dx
Q.13 1  cos x
is equal to: [IIT-1999] x
/ 4 Q.19 Let f : (0, )  R and F(x) =  f (t) dt .
1 1 0
(A) 2 (B) –2 (C) (D) –
2 2 If F(x2) = x2 (1 + x). Then f(4) equals-
[A] [IIT-2001]
(A) 5/4 (B) 7 (C) 4 (D) 2 [C]

cos 2 x
Q.14  1 a x
dx, a > 0 [IIT Scr. 1/ 2
  1 x 
 Q.20 The integral   [ x ]  n    dx
1 x 
2000]  (1 / 2 ) 

(A)  (B) a equals -

(C) 2 (D) 2 [C] [IIT Scr. 2002]
(A) – 1/2 (B) 0
x (C) 1 (D) 2 n(1/2) [A]
Q.15 Let g(x) =  f (t) dt where t is such that
0 x
Q.21 Let F(x) =  2  t 2 dt , then the real roots of
1 1
f(t) 1 for t  [0, 1] and 0  f(t)  for 1
2 2
the equation x2 – F(x) = 0 are- [IIT-2002]
1
t [1, 2]. Then g(2) satisfies the inequality: (A) ± 1 (B) ±
2
[IIT-2000]

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 1 1
(C) ± (D) 0 and 1 [A] 1 x
2 Q.26  1 x
dx equal to -
0
[IIT Scr.2004]
Q.22 Let T > 0 be a fixed number. Suppose f is a
 
continuous function such that for all x  R; (A) + 1 (B) –1 (C) 1 (D) 
2 2
T
[B]
f(x + T) = f(x) .If I =  f ( x ) dx , then the
0 0

3 3T
Q.27  [x3 + 3x2 + 3x + 3+ (x + 1) cos (x + 1) dx =
2
value of  f (2 x ) dx is-
[IIT Scr.2005]
3
[IIT-2002] (A) 4 (B) 0 (C) –1 (D) 1 [A]
3 1
(A) I (B) I 
t
2
2 Q.28 f ( t ) dt  1  sin x ; 0 x  ,
sin x 2
(C) 3I (D) 6I [C]
 1 
1 then f   is -
Q.23 (m, n) =  t m (1 + t)n dt, then m, n =?  3
0 [IIT Scr.2005]
[IIT Scr. 2003] 1
(A) 3 (B) (C) 1 (D) 3 [A]
n I ( m 1, n 1) 3
(A) I ( m, n )  . Q.31 Let f be a non-negative function defined on the
m 1 m 1
1 I ( m 1, n 1)
(B) I ( m , n )  . x
m 1 m 1 interval [0, 1]. If  1  (f ( t )) 2 dt =
0
2n n.I ( m 1, n 1)
(C) I ( m , n )  
1 m m 1
x
2n n. I ( m 1, n 1)
 f ( t ) dt ,
(D) I ( m , n )  
1 m m 1 0

[C] 0  x  1, and f(0) = 0, then- [IIT- 2009]

x 2 1 1 1 1 1
2 (A) f   < and f   >
Q.24 If f(x) =  e  t dt . Then f(x) increase in- 2 2 3 3
x2
[IIT-2003] 1 1 1 1
(B) f   > and f   >
(A) (–2, 2) (B) No value of x 2 2 3 3
(C) (0, ) (D) (–, 0) [A] 1 1 1
1
(C) f   < and f   <
t2
2 2 3 3
2
Q.25 If  x f ( x ) dx = 5 t5 for t > 0, then f (4/25)
1 1 1 1
0 (D) f   > and f   < [C]
2
  2 3 3
is equal to - [IIT Scr.2004]
2 2
(A) – (B) 0 (C) (D) 1 [C]
5 5 1

Q.4  max. {x–[x], – x – [– x]} is equal to

1

(A) 3 (B) 3/2

(C) 3/4 (D) 2
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 1 
sin x 
Sol.[B] 
 2 max. [{x}, {– x}]dx Q.5 If Dirchlet's integral I = 0
x
dx =
2
0
 3
Q.5 The smallest interval in which value of sin x
then 
0
x
dx is –
1
xdx

0
x 3  16
lies (A) /2
(C) 1
(B) /4
(D) 0
Sol.[B]
 1 
(A) 0,  (B) [0, 1]
 17  sin t t
sin z 2
 1   1 Q.6 If x =  arc sinz dz, and y =
 z
dz
(C) 0,  (D) 0,  C2 n
 27   7 dy
1 1 1 then is –
xdx xdx xdx dx
Sol.[A] x 2
 16
 x 3
 16
 x 4
 16 tan t tan t
0 0 0 (A) (B)
2t t2
/ 2
cos 3x  1
tan t tan t 2
Q.1 
0
2 cos x  1
dx is – (C)
2t 2
(D)
2t 2
Sol.[C]
(A) 2 (B) 1
(C) 1/2 (D) 0 3 4

Sol.[B] Q.7
/ 4
 1  sin 
d is –
1/ 2 1/ 2
 x  1  2  x  1  2  (A)  ( 2  1 ) (B)  ( 2  1 )
Q.2    
 x  1 
1 / 2 
 x 1
  2

dx is –
(C)  ( 2  2 ) (D)  ( 1  2 )
(A) 4n 4/3 (B) 2n 4/3 Sol.[A]
(C) –2n 4/3 (D) –4n 4/3
Sol.[A]

x
e t Q.8  (cosax – sinbx)2 dx where a and b are
Q.3 If f(x) = 
0
t
dt ; x > 0 and 
integers, is–
x (A) – (B) 0
et

1
ta
dt  e a [f ( x  a )  f (k )] then
Sol.[D]
(C)  (D) 2

possible value of k is –
1 / 4
(A) a (B) 1 – a 1 1
(C) 1 + a (D) a – 1 Q.9 If 0
x
cot 1  dx =
x  f ( x ) . cosec 2x
0
Sol.[C]
dx then f (x) =
1/ 2 (A) x (B) 2x

Q.4 If 0 < x <
2
then  cot x d (cosx) =
Sol.[B]
(C) 0 (D) x/2
3/2

1 2 1 3 Q.10 If
(A) (B)
2 2 1 x 1 x 1 x
f (x)  tan  tan  ..... n tan n  ...
1 3 1 2 2 2 4 4 2 2
(C) (D)
2 2 /2
Sol.[B] then  f ( x )dx has the value –
 6
(A) n2/3 (B) n3/2
(C) n3/ (D) n2/
Sol.[B]

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 x Q. 10 The number of values of x satisfying
Q.11 Let f : (0, )  R and g(x) =  0
t f(t) dt. If
2[ x 14]  x  {x}

g(x2) = x4 + x5, then

12

 f (r 2
) is –
 0
 dx 
2 0 
[ x  14] dx , is

r 1 equal to (where [ . ] and {.} denote the greatest

(A) 216 (B) 219 integer and fractional part of x.)
(C) 221 (D) 233 (A) 11 (B) 12
Sol.[B] (C) 13 (D) 14
x 1 Sol.[D]
Q.12 If 
0
f ( t ) dt = x +  x
t f ( t ) dt , then
the value of f(1) is 2

(A) 1/2 (B) 0 Q. 11  | (1 – x )n x | dx is equal to

1 0
(C) 1 (D) – (A) – 1 (B) 0
2
Sol.[A] (C) 1 (D) 2
Q.13 A function F is defined by the following Sol.[C]
100
x et Q. 12 Evaluate :  [tan –1 x ] dx ; where [ ]
definite integral F(x) =
1 t
dt if x >  0

represents greatest integer function.

x e t (A) 100 – tan 1 (B) 100.tan 1
0.Then integral
 1 t2
dt can be expressed as
(C) 25  (D) 100 –

(A) F(x) – F(e) (B) F(x) – e + ex 4
Sol.[A]
ex e x 3 4
(C) F(x) + e– (D) F(x) – +e 
Sol.[C]
x x Q.18 
/ 4
1  sin 
d is –

2x  1 (A)  ( 2  1 ) (B)  ( 2  1 )
Q.1 If x 4
 2x 3  x 2  1
dx = A n
(C)  ( 2  2 ) (D)  ( 1  2 )
Sol.[A]
/2
x 2  x 1
 C then the value of A can be
Q.5 If I10 =  0
x 10 sin x dx , then I10 + 90 I8 is
1 x2  x
(A) 10(/2)6 (B) 10(/2)9
(A) 1 (B) 1/2 (C) 10(/2)8 (D) 10(/2)7
(C) –1/2 (D) –1 Sol.[B]
2 n 1 / 2
Sol.[C] x 
 dx
Q.6 The value of  (sin x )  2  dx is
Q. 8 The value of  0 1  2 tan x
is – 2n

(where {x} denotes the fractional part of x}-


(A) 0 (B) 2 n   1 n
4 (A) (B)
 2 

(C) (D)  ( n  1)
2 (C) (D)
Sol.[C] 
2 n  1
2 / 5
sin x 2
Q. 9 The value of 
 / 10
sin x  cos x
dx , is equal
2 n 1 / 2
x 
to Sol.[A] I =  (sin x )   dx
2
2n
(A) /2 (B) 3/2
(C) 5/2 (D) none of these
Sol.[D]

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 2 (A) I1 > I3 > I2 (B) I3 > I1 > I2
x
= 2n  (sin x ) 2
dx + (C) I1 > I2 > I3 (D) I3 > I2 > I1
0 Sol.[A] sinx < x  x  (0, )
so, cos(sin x) > cosx, so I1 > I3
1/ 2
x and sin sinx > sinx
 (sin x ) 2
dx / 2 / 2
0 so 
0
sin(sin x )dx >  (sin x )dx
0
2n  1 / 2 /2
=
2
sin t t 2
 sin(cos x )dx >  (cos x )dx I
0 0
2 >

 arc sin z dz, and y = 

sin z
Q.7 If x = dz I3
z
C2 n  1
 (3x
2
dy Q. 1 Evaluate sin  x cos )dx
then is – x x
dx  
tan t tan t (A) x sin +C (B) x2 sin + C
(A) (B) x x
2t t2
tan t tan t 2 
(C) (D) (C) sin +C (D) none of these
2t 2 2t 2 x
Sol.[C] Sol. (D)
x nx
Q.2 Let f be integrable over [0, a] for any real a. If dt et
Q. 2 If Li(x) = 
2
n t
, if x > 2, and let 
b
t
dt
/2

we define I1 = 
0
cos  f(sin  + cos2) d = Li(x) then the value of b is –
(A) e2
(B) 2
and
/2 (C) n2 (D) –n 2
I2 =  sin 2 f(sin  + cos2)d, then Sol. (C)
3
0 1 1 
(A) I1 = I2 (B) I1 = –I2 Q.6 If I = x 

sin   x 
x  dx, then I equals :
1/ 3
(C) I1 = 2I2 (D) I1 = –2I2 (A) 3 /2 (B)  + 3 /2
 (C) 0 (D) None of these
2 1
Sol.[A] I1 =
 cos  f(sin + cos2) d
0
Sol. [C] Put x =
t
1/ 3
  1  1 
2 I=  t sin  t  t    t 2  dt
 2 sin  cos  f(sin + cos ) d
2
I2 = 3
1/ 3
0 1  1
1 =  sin   t  dt
 t  t
 f (t  1 – t
2
Let sin = t, then I1 = ) dt. 3

0 =–II=0
1

 t f (t  1 – t Q.5 Let ƒ(x) and g(x) be two functions satisfying

2
and I2 = 2 ) dt
0 ƒ(x2) + g(4 –x) = 4x3, g(4 – x) + g(x) = 0, then
replace t by 1 – t then I1 = I2 4
/ 2 /2 the value of 4
ƒ( x 2 ) dx is-
Q.3 If I1 =  cos (sin x) dx; I
0
2 =  sin (cos
0
(A) 512 (B) 64
x)dx and (C) 256 (D) 0
/ 2 4 4

I3 =  cos x dx, then - Sol.[A] I =  4

f ( x 2 )dx = 2 0 f ( x 2 )dx
0

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 4
= 2 0 {4 x 3  g ( 4  x )}dx 1 1
(A) (B) (C) 3 (D) 3
3 3
/2 3
sin x cos x
Q.8 Value of I =
 sin 4 x  cos 4 x
dx is equal 1

t
0 2
Sol.[C] f ( t )dt  1 – sin x {given}
to sin x

(A) /8 (B) /4 Differentiating both sides w.r.t. x using Newton

(C) /2 (D)  Leibnitz formula, we have

– sin2 x f (sin x) cos x = – cos x
/2 cos 3 x sin x
Sol.[B] I = 
sin 4 x  cos 4 x
0
dx sin2 x f (sin x) cos x = cos x {cos x  0}
/2 sin x cos x 1 
I+I=
0 
sin 4 x  cos 4 x
dx f(sin x) =
sin x2 , x  (2n + 1)
2
put sin2x = t 1
x f(x) =
x2
Q.14 Let f(x) be a function defined by ƒ(x) =  t (t 2

1
 1 
– 3t + 2)dt, 1  x  3. Then the range of f(x) is - Now, f   = 3.
 3
 1 
(A) [0, 2] (B)   , 4
 4  x

 1 
Q.5 If ƒ is a continuous function such that  ƒ( t )
(C)   , 2 (D) None of these 0
 4 
Sol.[C] f(x) = x(x 2 – 3x + 2) = x (x – 1) (x – 2). The dt   as |x|  , then for all k  R, equation
sign scheme for f(x) is as shown in figure.
 f(x)  0 in 1  x  2 and f(x)  0 x
in 2  x  3 2 2
kx +  ƒ( t ) dt – a = 0 (a > 0) has -
 f(x) is decreasing in [1, 2] and 0
increasing in [2, 3]
2 (A) all roots in (–, 0)
 x(x
2
 min. f(x) = f(2) =  3x  2) dx (B) all roots in (0, )
1
2 (C) odd number of roots in (–, 0) and odd
x4 1 
=  
 x3  x 2  = number of roots in (0, )
 4 4 1
max. f(x) = the greatest among [f(1), f(3)] (D) None of these
x
– + – + Sol.[C] Let g(x) = k x + 2 2
 f ( t )dt  a … (1)
0 1 2 0
1 x
 x(x
2
f(1) =  3x  2)dx  0 Since, a > 0 and  f ( t )dt   as x  ± 
1 0
3
(given)
 x(x
2
f(3) =  3x  2)dx  2
1 0
 1 
 max f(x) = 2, so the range =  ,2 .
 g(0) = 0 +  f ( t )dt  a = –a < 0
0
 4 
1
 1  and g() =  +  – a =  > 0
Q.3 If 
t 2 (f(t)) dt = (1 – sin x), then f 
 3
 is
also, g(–) =  +  – a =  > 0
sin x

-  g(x) is continuous in (–,)

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 Hence, g(x) = 0 has odd number of roots in 9
= .
(–, 0) and odd number of roots in (0, ). 2
x 2
1 1
Q.6 Let ƒ(x) =  t sin
t
dt . Then the number of Q.2 Evaluate :  sin (sin x) dx -
0  / 2

points of discontinuity of the function ƒ(x) in 2 – 2

(A) – (B)
the open interval (0, ) is - 4 8

(A) 0 (B) 1 (C) 2 (D) infinite 2

(C) (D) None of these
1 4
Sol.[A] Since, f(x) = x sin
x Sol.[B] The graph of ƒ(x) = sin–1 (sin x) is shown as in
Now, at all points in (0, ), f(x) has a definite figure.
finite value.
 f(x) is differentiable finitely in (0, )
As a finitely differentiable function is also
continuous
 f(x) is continuous in (0, ). Y
1
Q.1 Evaluate :  max {2 – x, 2, 1 + x} dx. - /2
1
 3/2 2
7 9 10 9 X
(A) (B) (C) (D) –/2 /2
2 2 3 4
Sol.[B] The graph of ƒ(x) = max {2 – x, 2, 1 + x} is –/2
shown as in figure.
2 0
Therefore  sin 1 (sin x) dx =  sin
1

 / 2  / 2

(sin x) dx
Y  2
+  sin 1 (sin x) dx +  sin
1
(sin x) dx
3 ƒ(x) = max {2, –x, 2, 2 + x} 0 

2 = Area of shaded region

1   1 
= –     +     –
1 2 2 2  2 2
X
–1 O 1
1 
1    
2 2
Therefore  max {2 – x, 2, 1 + x} dx = Area
1 2
=– .
of shaded region 8

1 2
= (2 + 3) × 1 + 1 × 2
2 Q.3 Evaluate :  max {x + |x|, x – [x]} dx, where
2

[x] denotes the greatest integer  x -

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 (A) 4 (B) 3 (C) 5 (D) 1 Put sin x – cos x = t then (cos x + sin x) dx = d t
2 and (sin x – cos x)2 = t2
Sol.[C]  max {x + |x|, x – [x]} dx
 1 – sin 2x = t2  sin 2x = 1 – t2
2

0 2 x0  t  –1
When 
=  max {0, x – [x]} dx +  max {2x, x – x  /4  t  0 
2 0
0
dt
[x]}dx Then I =  9  16(1  t 2 )
–1
the graph of ƒ(x) = max {x + |x|, x – [x]} is
0
shown as in figure dt 1
=  25  16 t 2
=
16
–1

0
dt
  5 2
1 2
Y   t
4 4
0
 5 
1 1 1  t 
.  4 
= 16 5 ln 
2. 5
X 4 

t 

–2 –1 O 1 2  4  1

1  1
= ln 1  ln 
40  9
2
therefore  max {x + |x|, x – [x]} dx = Area =
1
{0 + ln 9} =
1
2 ln 3
–2 40 40
of shaded region 1
= ln 3.
1  1 20
= 2   1  1 + ×2×4
 2  2 /3

=1+4 Q.5 Evaluate :  x cosec x dx -

2

/4
= 5.
 1 3
/4 (A) (3 3 – 4) + ln  
(sin x  cos x ) 12 3 2 2
Q.4 Evaluate :  (9  16 sin 2 x )
dx -
0
 1 2
(B) (3 3 – 4) + ln  
1 1 12 3 2 3
(A) ln 2 (B) ln
20 20
 1 3
4 (C) (3 3 – 2) + ln  
12 3 2 2
1
(C) ln 3 (D) None of (D) None of these
20
/3
these Sol.[A] Let I =  x cosec 2
x dx
/4 /4
(sin x  cos x )
Sol.[C] Let I =  (9  16 sin 2 x )
dx Integrating by parts taking x as first and cosec2 x
0
as second function, we have
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 /3 (C) (–2, 2) (D) No where

/3
I = x (– cot x) + cot x dx
/4
x 2 1
/4 2
Sol.[B] ƒ(x) =  e – t dt
 1  /3 x2
= –  .   + ln | sin x |
3 3 4 / 4 e – ( x 2 1) 2  e – ( x 2 )2 
 
 ƒ(x) = 2x  
   3 1 
=– + +  ln  ln
3 3 4  2 2  4
 1 2 x 2 ) 2
1
= 2x e – ( x [1 – e 2 x ].
 1 3 Now ƒ(x) > 0 for x  (– , 0).
= (3 3 – 4) + ln   .
12 3 2 2 1

t
m
Q.6 Let ƒ(x) be a function satisfying ƒ(x) = ƒ(x) Q.8 If I (m, n) = (1 + t)n dt ; m, n  R, then
0
with ƒ(0) = 1 and g(1) and g(x) be a function
I (m, n) is -
that satisfies ƒ(x) + g(x) = x 2. Then the value of
n
(A) I [(m + 1), (n – 1)]
1 1 m
the integral  ƒ (x) g(x) dx is - m
0 (B) I [(m + 1), (n – 1)]
n 1
e2 3 e2 3
(A) e +  (B) e – – 2n n
2 2 2 2 (C) – I [(m + 1, n – 1)]
1 m 1 m
e2 5 e2 5
(C) e + + (D) e – – 2n m
2 2 2 2 (D) – I [(m + 1, n – 1)]
1 m 1 n
1 1

 e
x 1 1
Sol.[B] ƒ( x ) g(x) dx = (x2 – ex) dx
 t m (1 + t)n dt =  (1  t )
n m
Sol.[C] I (m, n) = t dt
0 0
0 0
[ƒ(x) = ex satisfies ƒ (x) = ƒ(x) and ƒ(0) = 1]
[Integrating by Parts]
1 1
1
x exdx –  e
2 2x
= dx 2n n n 1
0 0
=
m 1
–
m 1
 (1  t ) tm
0
1 1

   e 2x  +1
dt
 (2 x ) e
1
= x 2e x 0 – x
dx –  
 2 
 
2n
0 0
n
= – I [(m + 1), (n –
 1  m 1 m 1
= e – 2  xe

x
  – e 1
0
x
dx  –

1 2 1
2
e +
2 1)].
 0 

2 x (1  sin x )
= e – 2 [e – (e – 1)] –
1 2 1
e +
Q.9  1  cos 2 x
dx is -
2 2 –

1 2 1 e2 3 (A) 2/4 (B) 2 (C) zero (D) /2

=e–2– e + =e– – .
2 2 2 2 
2 x (1  sin x )
x 2 1
Sol.[B]  1  cos 2 x
dx
2 –
Q.7 If ƒ(x) =  e – t dt, find the interval in which
 
x2 2x x sin x
ƒ(x) is increasing -
=  1  cos
–
2
x
dx  2  1  cos 2 x
–
dx

(A) (0, ) (B) (–, 0)

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   1 
x sin x x sin x x log x x log x
=0+4  1  cos 2 x
0
dx  4  1  cos 2 x
0
Sol.[B] I =  (1  x
0
2 2
)
dx +  (1  x
1
2
)2
dx

dx …(1) = I1 + I2.
 1
x sin x Put x = in I2 and adjust the limits
Now I = 2
dx = t
0 1  cos x
1 1  1
log .  dt
t  t 2 
0 1
 t  t log t

(   x ) sin(   x )
2 dx I2 =   1 
2 =  (1  t 2 ) 2
1  cos (   x ) 1 0
0 1  

 t2 
(   x ) sin x
=  1  cos 2 x
dx dt
0
1
x log x

 sin x

x sin x I2 = –  (1  x 2 ) 2 dx = –I1
I=  1  cos
0
2
x
dx –  1  cos 2 x
0
dx 0

Hence I1 + I2 = 0.

sin x 2
 2I =   dx [ x 2 1]
0 1  cos x
2 Q.11 The value of x dx, where [x] is the
0

 sin x greatest integer less than or equal to x is -

2
 1  cos 2 x dx (A) 2 (B) 8/3
0
(C) 4 (D) None of these
 From (1),
 
Sol.[D] For x  [0, 2], x + 1  [1, 5], we must break
2

2 x (1  sin x )  sin x

 2
1  cos x
dx = 4  
2
 1  cos 2 x dx [0, 2] = [0, 1]  [1, 2 ]=
– 0
[0, 1]  [1, 2 ] [ 2 , 3 ]  [ 3 , 2].

sin x
= 2  1  cos 2 x dx 2
0
[ x 2 1]
Put cos x = t so that – sin x dx = dt
Hence x dx
0
i.e. sinx dx = – dt. 1 2 3
[ x 2 1] 2 2

When x = 0, t = 1 ; when x = , t = –1 = x dx +  x [ x 1] dx +  x [ x 1]

0 1 2
 –1
2 x (1  sin x )  dt
  2 dx = 2  1 t2 dx
  1  cos x 1 2
[ x 2 1]
x
= –2 tan  1
t  –1
1
+
3
dx

= –2 [tan–1(–1) – tan–1 (1)] 1 2 3 2

x
4
=  x dx   x 2 dx   x dx +
3 dx
   0 1 3
= –2  –   = 2. 2
 4 4
1 1 3/2 1 1
 = + [2 – 1] + [9 – 4] + [32 –
x log x 2 3 4 5
Q.10 The value of integral  (1  x 2 ) 2 dx is -
0 35/2]
(A) 1 (B) 0 469 1 3/2 1 5/2
= + 2 – 3 .
(C) 2 (D) None of these 60 3 5
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 3 1 m ( m  2)
Q.12 The value of  {| x  2 | [ x ]} dx, where [x] (C)
1 m
,
m 1
(D)
–1
denotes the greatest integer less than or equal to
x is - m
,m–2
(A) 5 (B) 7 (C) 4 (D) 3 m 1
3 0

  {| x  2 | [ x ]} dx
e
Sol.[B] {| x  2 | [ x ]} dx =
 (log x )
m e

–1 –1
Sol.[A] Im = dx = x (log x ) m 1
1
1
e
+  {| x  2 | [x ]}dx + – m  (log x )
m 1
dx
0
1
2 3
= e – m
 
{| x  2 | [ x ]}dx + {| x  2 | [ x ]}dx
1 2

0  e e 
 x (log x ) m 1  ( m  1)  (log x ) m  2 dx 
=  (2 – x  1)dx +  1 
–1  1 
1 2 = e – me + m (m – 1) Im–2 = (1 – m) e + m (m – 1) Im–2
 ( 2 – x  0)dx +  (2 – x  1)dx + Im
0 1 So + m Im– 2 = e. Thus K = 1 – m and L =
1 m
3

 ( x – 2  2)dx 1
.
2 m
0 1 2
x2 x2 x2 16
= x + 2x  + 3x – + 1
2
–1
2
0
2
1
Q.14 The value of  tan x 1 dx is -
1

x2
3 16 4
(A) +2 3 (B) –2 3
2 3 3
2

 1  1 4 16
= –   1   +  2   + (6 – 2) – (C) +2 3 (D) –2 3
2 2 3 3
 
Sol.[D] Integrating by parts, the given integral is
 1 equal to
3  
 2 16
x 1

16
x tan–1 x 1 –
9 1
x 4 x x 1
+ – 2 = 7. 1
2
dx
e
I m I m2 16
Given Im =  (log x )
m
Q.13 dx. If  =e 16 1 dx
1 K L =
3
–
4
 x 1
1
then values of K and L are -
3
16 1 4 t (1  t 2 )
(A) 1 – m,
1
(B)
1
,m
=
3
 –
4  t
dt ( x = 1 +
m 1 m 0

t 2)
16 16
= –( 3 + 3 )= –2 3 .
3 3
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  /2
Q.15 The value of  [2 sin x ] dx, where [ . ] = 2n  (sin x  cos x )dx +
0 0

represents the greatest integer function is -

t
(A)  (B) 2 (C) – (D) 2/3

 (sin x  cos x )dx
0
Sol.[D]  [2 sin x ] dx = (2n) (2) + sin t – cos t + 1 = (4n + 1) + sin t – cos t.
0
3 / 4
/6 /2 dx
=  [ 2 sin x ]dx +  [ 2 sin x ]dx
Q.2  1  cos x
is equal to -
/ 4
0 /6
(A) 2 (B) –2 (C) 1/2 (D) –1/2
5 / 6
+  [2 sin x ]dx + 3 / 4
dx
3 / 4
dx
/ 2 Sol.[A] I =  1  cos x
=  1  cos x (  x )
/ 4 /4

 (Property 8)
 [ 2 sin x ]dx 3 / 4
dx
5 / 6
=  1 – cos x
/6 /2 5 / 6 / 4
=  0 dx +  1 dx +  1 dx + 3 / 4
 1 1 
0 /6 /2
2I =     dx =
 1  cos x 1  cos x 
/4

3 / 4
 0 dx 
2
dx
5 / 6
/4 1  cos 2 x
  5  2 3 / 4
= – + – = . = – 2 cot x  / 4 = 4. Hence I = 2.
2 6 6 2 3
Q.3 Let ƒ be a positive function and
Q.1 The value of the integral k
n  t I1 =  x ƒ( x (1  x )) dx, I2 =
 (| cos x |  | sin x |) dx is - 1 k
0 k

(A) n (B) 2n + sin t + cos t  ƒ( x (1  x )) dx where 2k – 1 > 0 then

1 k
(C) cos t (D) sin t – cos t + 4n + 1 I1/I2 is -
(A) 2 (B) k (C) 1/2 (D) 1
Sol.[D] Since the period of |sin x| + |cos x| is /2 so b b
n  t Sol.[C] Since  ƒ( x ) dx =  ƒ(a  b  x ) dx, we
 (| cos x |  | sin x |)dx = a a
0 have
k
/ 2
2n  (| sin x |  | cos x |)dx +
I1 =  (k  1  k  x ) ƒ ((k + 1 – k – x) (1
1 k
0
–(k + 1 – k – x)) dx
t k

 (| sin x |  | cos x |)dx =  (1  x ) ƒ((1 – x)x) dx

0 1 k

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 k k 
sin x
=  ƒ((1  x )) dx –  x ƒ((1  x ) x ) 2I = 4  2 dx
1 k 1 k 0 1  cos x

dx = I2 – I1. Let cos x = t

So 2I1 = I2 and I1/I2 = 1/2.  sin x dx = – dt
1/ 2 –1 1
 1 x    dt  dt
Q.4 The integral   [ x ]  log

 dx equals
1 x  I = 2   2
 = 2 
2 = 2 [tan
–1

1 / 2 1 1 t  11  t

-
t ]11
(A) –1/2 (B) 0
(C) 1 (D) 2 log (1/2)  
= 2 [tan–1 (1) – tan–1 (–1)] = 2   
1/ 2 1/ 2
4 4
 1 x 
Sol.[A] 
 [ x ]  log

 dx =
1 x 
[ x ]dx  = 2 ×

= 2.
1 / 2 1 / 2
2
1 x Q.6 The value of the integral
(since log is an odd function)
1 x
1/ 2 2 2
 x 1  x 1 
0 1/ 2 0
  x  1    x  1   2 dx is -
=  [x] dx +  [ x ] dx =  (–1) dx 1 / 2
1 / 2 0 1 / 2
4 4
(A) 2 log (B) 4 log
1 3 3
=– .
2 4
(C) log (D) None of these
 3
2 x (1  sin x )
Q.5  1  cos 2 x
dx is - 1/ 2
 x 1 x 1
2
 Sol.[B]    
 x 1 x  1
1 / 2
2 
(A) (B) 2 (C) zero (D)
2 1/ 2
4 x 1 x 1

=  
x 1 x 1
dx
2 x (1  sin x ) 1 / 2
Sol.[B]  1  cos x 2 dx
1/ 2 1/ 2
– 4x 4x
 
=  x 12
dx  2  2
x 1
dx
2x x sin x 1 / 2 0
=  1  cos 2 x dx + 2  1  cos 2 x dx
(integrand is an even function)
– –

 1/ 2
 4x 
I=0+4
x sin x dx
 1  cos 2 x = –2   2 dx
0 0  x 1 


x sin x dx  4x  1 
  0 in the interval  0,  
I=4  1  cos 2 x  x2 1  2 
0

  3
(   x ) sin x = –4[log (1 – x2) ]10/ 2 = – 4  log  = 4 log
I=4  1  cos 2 x
 4
0

 
sin x x sin x 4
 I = 4  1  cos 2 x dx – 4  1  cos 2 x  .
3
0 0

dx
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 n2 2[ x ] 2[ x ].1
Q.7 [ x ] dx is equal to - Sol.[A]  ( x  [ x ])dx =  ( x  [ x ])dx
0 0 0

1
n ( n  1)( 4n  1)
(A)
6
(B) = 2[x]  ( x  [ x ])dx
0

[x – [x] is a periodic function of period 1]

n ( n  1)( 4n  1)
6  2 
1 1 
 x
= 2[x]     [ x ]dx  = 2 [x]  1 
 2   

  0 =
2 
n ( n  1)( 4n – 1)  0 0 
(C) (D) None of
6
[x].
these
n2
   
1 4
Sol.[B]   x  dx = x dx + x dx +
0 0 1

Q.9 For x  R and a continuous function ƒ, let

 
9
1 cos 2 t
x dx
4
I1 =  x ƒ {(2 – x)} dx and
sin 2 t
n2

  1 cos 2 t
+ …. + x dx
( n 1) 2
I2 = 2
ƒ{x(2 – x)} dx. Then I1/I2 is -
sin t

(A) 0 (B) 1 (C) 2 (D) 3

  x   0, if 0  x  1;1, if 1  x  4;  1 cos 2 t


 


2 2
2, if 4  x  9.....; ( n  1), if ( n  1)  x  n  Sol.[B] I1 = x ƒ {(2 – x)} dx
sin 2 t
1 4 9
=  0dx +  1 dx +  2 dx + …. 1 cos 2 t
0 1 4 = I1 =  (2 – x) ƒ(x(2 – x)) dx = 2 . I2 –
sin 2 t
n2
+  (n  1) dx I1
( n 1) 2
I1
= 1(4 – 1) + 2(9 – 4) + ….. + (n – 1) [n2 – (n – 1)2]  2I1 = 2I2  = 1.
I2
= – (12 + 22 + 32 + …. + n2) + n3

x sin x
= n3 –
n ( n  1) ( 2n  1)
=
n ( n  1) ( 4n  1)
.
Q.10 Evaluate :  1  cos 2 x dx
6 6 0

2[ x ] 2 2
(A) (B)
Q.8 The value of the integral  ( x  [ x ])dx is 2 8
0
2
- (C) (D) None of these
4
1
(A) [x] (B) [x] 
x sin x
2 Sol.[C] Let I =  1  cos 2 x dx … (1)
(C) 3[x] (D) 2[x] 0

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  1
(   x ) sin(   x )
=  2
(By Prop. =  (1  x ) x99 dx
0 1  cos (   x ) 0
1
IV)
 (x
99
=  x100 ) dx
 0
(   x ) sin x
=  1  cos 2 x
… (2)
=
x100

x101
1

0 100 101
0
Adding (1) and (2), we get 1 1
= –

sin x
100 101
2I =   1  cos 2 x dx =
1
.
0
10100
Put cos x = t – / 6
When x = 0  t = 1 dx
x =   t = –1
Q.12 Evaluate :  tan 2 n x
 / 3 1 
–1
dt
 2I = –  1 t2 (A)

(B)

1 6 12
1
dt 
=  1 t2 (C)
3
(D) None of these
–1
 / 6
=  tan–1 t
1 dx
1 Sol.[B] Let I =  tan 2n x
 / 3 1 
=  {tan 1 – tan (–1)}
–1 –1

/3
=  {2 tan–1 1} dx
=  (tan (– x )) 2 n
(By Prop.
 /6 1 
 I =  tan–1 1 =  .
4 X)
/3
2 cos 2n x dx
Hence I =
4
. =  (sin 2 n x  cos 2n x )
… (1)
/6
1
and
Q.11 Evaluate :  x (1 – x) 99
dx
0 I =
1
(A) (B)
10100
  
/3 cos 2n    x dx
1 3 6 
11000  2 n     2n    
 / 6 sin    x   cos    x 
1 3 6  3 6 
(C) (D) None of
10010 (By Prop. V)
these /3
sin 2n xdx
1 =  cos 2n x  sin 2 n x
… (2)
Sol.[A] Let I =  x (1 – x)99 dx / 6

0 Adding (1) & (2) we get

1 /3
  
=  (1  x ) (1 – (1 – x))99 dx (By Prop. IV) 2I =  1 . dx =
3
–
6
=
6
0 /6

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  It is clear from (1), (2) and (3) we get
I= .
12 b
|x|
199   x
dx = |b| – |a|.
 1  cos 2 x 

0
Q.13 Evaluate :   dx -
 2 
– 1
sin t
(A) 200 (B) 400 (C) –200 (D) –400 Q.1 Given  1 t dt = , find the value of
0
199 4
 1  cos 2x  sin( t / 2)
Sol.[B] Let I =  
 2
 dx
  4  2  t
in terms of  -
– 4 2
199  (A)  (B) – 
=  | sin x | dx (C) 2 (D) None of these
 Sol.[B] Let
 4
sin( t / 2)
= (199 – (–1))  | sin x | dx I=  4  2  t
dt
0 4 2
( | sin x | is periodic with period ) = (4 – (4 – 2))

 ( 4  (4  2)) t  4  2 
= 200  sin x dx 1 sin 
 2

 dt
0

 4  2  ((4  (4  2) t  4  2)
= 200 (– cos x) 0 0
= 200 (1 – (–1)) = 400. 1
sin( t  1)
b
=2  (4  2t )
dt
|x| 0
Q.14 Evaluate :  x
dx, a < b - 1
sin( t  1)

a
= dt
(2  t )
(A) |a| – |b| (B) |b| + |a| 0
1
(C) |b| – |a| (D) None of these sin(1  t  1)
b
=  2  (1  t )
0
|x|
Sol.[C] Let I =  x
dx (By Prop. IV)
a 1
sin(  t )
Case I : When 0 < a < b
=  (1  t )
dt
0
b b 1
x sin t
   1 t
b
then I = dx = 1 . dx = x a =b–a =– dt
x
a a 0

= |b| – |a| … (1) = – . (given).

16 
Case II : When a < 0 < b 3
Q.2 Evaluate : dx
0 b  sin x
1 1
0
then I = – . dx + . dx
a 0 21 21
(A) (B)
4 2
= – (0 – a) + (b – 0)
11 11
=b+a (C) (D)
2 4
= |b| – |a| … (2) 16 
3
Sol.[B] Let I =
Case III : When a < b < 0
 sin x
b 0
then I =  ( 1) . dx = – (b – a) = – b + a 5 

3
0 = dx
= |b| – |a| … (3)
 | sin x |
0

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   
5 5 
3 (A) – ln 2 (B) – ln 4
=  | sin x | dx +
 | sin x | dx 2 2
0
5 
(C) – ln 2 (D) None of these
 /3 4
= 5  | sin x | dx +  | sin x | dx 1
1
0 0 Sol.[A] Let I =  cot (1 – x + x2) dx
(By Prop. XIII & XV) 0
( | sin x| is periodic with period ) 1
1
 /3 =  cot (1 – x (1 – x)) dx
=5  sin x dx +  sin x dx 0
0 0 (0  x < 1)
 /3 1
= 5 (– cos x)  (  cos x )
1  1 
1  21
0 0
=  tan   dx
 1  x (1  x ) 
0
= 5 (1 + 1) –  1 = .
 2  2 1
 x  (1  x ) 
1
1
ln x
1
ln (1  x )
=  tan   dx
 1  x (1  x ) 
Q.3 Show that  (1  x ) dx = –  x
dx 0
1
0 0
1
2 =  (tan x  tan 1 (1  x )) dx

=– 0
12 =
2 2 1 1
(A) (B) 1 1
12 6  tan x dx   tan (1  (1  x ))dx
0 0
2 2 (From property)
(C) – (D) –
6 12 1
1
1
ln x = 2  tan x dx
Sol.[D] Let I =  (1  x )
dx 0
0 Integrating by parts taking unity as the second
Integrating by parts taking ln x as first function, function, we have
we have  1 
1 x

1 1
ln (1  x ) I = 2 [tan x . x ]0  dx 
=  lnx.ln (1  x ) 0 – 
1 2
dx  0 1 x

x  
0
 1 2 1
1
ln (1  x ) = 2   [ln | 1  x ]0 
I=0–  x
dx 4 2 
0  1 
 2 3 4  = 2   ln 2 
1  x  x  x  x  ....  4 2 
 
= –  2 3 4
 dx 
0 x  Hence I = – ln 2.
  2
1 1
x x2 x3 
= – 1  
 2 3

4
 ....  dx

Q.5 Evaluate :  [ x[1  sin x ]  1] dx, where
0  1
1 [.] is the greatest integer function -
 x2 x3 x4 
= –  x  2  2  2  ... (A) 2 (B) 1
 2 3 4  0 (C) 3 (D) None of these
1
 1 1 1 
= – 1  2  2  2  ...  Sol.[A] Let I =  [ x[1  sin x ]  1] dx,
 2 3 4  1
2

=– .
12
1
1
Q.4 Evaluate :  cot (1 – x + x2) dx -
0
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 0 2
 1 3
= a2     – a (0 – 1) – 20 = 2a + a –
=  [ x[1  sin x ]  1] dx +
 12 4  3
1
1 20.
Thus the given inequality is (2a2/3) + a – 20  – a2/3
 [ x[1  sin x ]  1] dx i.e. a2 + a – 20  0  – 5  a  4
0
Now [1 + sin x] = 0 if –1 < x < 0 Since a is a positive integer so a = 1, 2, 3, 4.
x
and [1 + sin x] = 1 if 0 < x < 1
 cos t
2
dt
0 1 Q.8 The value of xlim
0
is -
I=  1 . dx +  [ x  1] dx 0
x
–1 0
1 (A) 1 (B) 0
=1+1  dx (C) – 1
x
(D) 2
0
 cos t
2
=1+1=2 Sol.[A] Let ƒ(x) = dt and g(x) = x. Then
2 0
[ x 2 1]
Q.6 The value of x dx, where [x] is the ƒ(0) = g(0) = 0.
0
 xlim
ƒ( x ) lim ƒ( x )
greatest integer less than or equal to x is - 0 g ( x ) = x 0 g( x )
(A) 2 (B) 8/3
x
(C) 4 (D) None of these
 cos t
2
lim dt
Sol.[D] For x  [0, 2], x2 + 1  [1, 5], we must break  x 0 0 = xlim
0
[0, 2] = [0, 1]  [1, 2 ]  [ 2 , 3 ]  [
2 x
2
 1] 2
3 , 2]. Hence  x
[x
dx cos x .1  cos 0.0
0 1
1 2
[ x 2  1] cos x 2
x = xlim
2
 1]
= dx +  x[x dx +
0
= cos 0 = 1.
0 1 1
3
2
Q.9 The least value of the function
 x [ x  1] dx x
2 F(x) =  (3 sin u  4 cos u ) du on the
2 0
2
[ x  1]
+ x dx interval (5/4, 4/3] is -
3
54 3
1 1 1 1 (A) 3/2 – 3 /2 (B)
= + [23/2 – 1] + [9 – 4] + [32 – 2
2 3 4 5
74 3 94 3
35/2] (C) (D)
2 2
469 1 3/2 1 5 / 2
= + 2 – 3 . Sol.[D] We have F(x) = 3 sin x + 4 cos x. Since sin x
60 3 5 and cos x assume negative values in the third
Q.7 If a is a positive integer, then the number of quadrant, we have F(x) < 0 for all x  (5/4,
values of a satisfying 4/3) so F(x) assumes the least value at the
point x = 4/3. Thus the least value is
/ 2
  cos 3x 3   4 / 3
 a 
  4
 cos x   a sin x  20 cos x 
4   F(4/3) =  (3 sin u  4 cos u )du
0
0
a2 4 / 3
dx  are - = (–3 cos u + 4 sin u ) 0
3
4 4
(A) one (B) two = –3 cos + 4 sin – (–3)
(C) three (D) four 3 3
Sol.[D] The L.H.S. of the above inequality is equal to 9 4 3 94 3
= – = .
 sin 3x 3  2 2 2
a2   sin x  – a cos x – 20
 12 4 
/2
sin x 0

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Q.10 The difference between the greatest and the 1 cos2 t
x Sol.[B] I1 =  (2  x )ƒ((2  x )(2  (2  x )))
least value of the function F(x) =  ( t  1) dt sin 2 t
0 dx
on [2, 3] is - 1 cos 2 t
(A) 3 (B) 2
(C) 7/2 (D) 3/2
=  (2  x )ƒ(x (2  x )) dx
sin 2 t
Sol.[C] Differentiating the given function, we get
1 cos 2 t
dx d0
F(x) = [t + 1]t = x
dx
– [t + 1]t = 0
dx
=x+ = 2  ƒ( x (2  x )) dx
2
–
sin t
1.
1 cos2 t
This is positive for all x  [2, 3], so F is an
increasing function in this interval. Therefore its  x ƒ( x (2  x )) dx
3 sin 2 t
greatest value is F(3) =  ( t  1) dt and its = 2I2 – I1
Therefore, 2I1 = 2I2 and so I1/I2 = 1.
0
2 102
1
least value is F(2) =  ( t  1) dt, so that the Q.13 Evaluate :  [tan x ] dx, where [·] denotes
0 0
required difference between these values is the greatest integer function less than or equal to
3 2 x-
7
 ( t  1) dt –  ( t  1) dt =
2
. (A) 102 – tan 2
(C) 101 – tan 2
(B) 102 – tan 1
(D) None of these
0 0
102
1 1
Q.11 The mean value of the function ƒ(x) =
x x2
Sol.[B]  [tan x ] dx
0
on the interval [1, 3/2] is - tan 1 102
(A) log (6/5) (B) 2 log (6/5) =  [tan 1 x ] dx +  [tan
1
x ]dx
(C) 4 (D) log 3/5 0 tan 1
b 102
1
Sol.[B] Mean value =
ba  ƒ( x )dx = 0 + 1.  1.dx
a tan 1
(Property 14) = 102 – tan 1.
3/ 2
1 1 n
=
3 / 2 1  2
x x
dx = 2
 [ x ]dx
1
0
3/ 2 Q.14 Evaluate : , where [x] and {x}
1 1  n
  x  x  1  dx
   {x}dx
1
0
3/ 2
= 2 (log x – log (x  1) 1 = 2 [log (3/2) – log denotes the integral part and fractional part
(5/2) – (log 1 – log 2)] function of x and n  N -
= 2 log (6/5). (A) n (B) n – 1
Q.12 For any t  R and ƒ a continuous function, let (C) 2n (D) None of these
1 cos 2 t n
I1 =  x ƒ( x (2  x ))dx  [ x ]dx
2
sin t 0
Sol.[B] Let I = n
1 cos 2 t
and I2 =  ƒ( x (2  x ))dx then I1/I2 is  {x}dx
0
sin 2 t
equal to-
(A) 2 (B) 1
(C) 4 (D) None of these

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 = e

 (log x )
m e
1 2 3 n Sol.[A] Im = dx = x (log x)m 1
– m
  
[ x ]dx  [ x ]dx  [ x ]dx  ...   [ x ]dx
e
1
0 1 2 n 1
m 1
1  (log x ) dx

n {x}dx 1
0 m = e –
2 3 n  e e 
 x (log x ) m 1  ( m  1) (log x ) m  2 dx  
 
0  1. dx  2 dx  ...  ( n  1)  dx  1 
1 2 n 1  1 
= 1 = e – me + m (m – 1) Im–2 = (1 – m) e + m (m – 1) Im–2

n . x dx
So
Im
+ m Im – 2 = e. Thus K = 1 – m
0
1 m
0  1  2  3  ...  ( n  1)
1
1 and L = .
n. m
2 1 1
( n  1) n Q.2 A function ƒ is defined by ƒ(x) = r 1
, <
2 2r
=
2 = (n – 1). 1
n x  r 1 , r = 1, 2, 3…. then the value of
2
2 1
Q.15 Find the mean value of x (a – x) over the range
(0, a) -
 ƒ( x ) dx is equal -
0
a2 a2 (A) 1/3 (B) 1/4 (C) 2/3 (D) 1/3
(A) (B)
3 6 1  2 ( r  1)
1
(C)
a
(D) None of these
Sol.[C]  ƒ( x ) dx =   2 r 1
dx =
0 r 1 2r
6

a 1
 2 r 1 [2 – (r – 1) – 2–r]
Sol.[B] Mean value =  x (a  x )dx r 1
0  
( a  0) =  2  2( r 1) –  2  2r 1
1 1
a
1 
 1
 (ax  x
2
=
a
) dx = (22 – 2)  2  2r =2. .
1  1/ 4
=
0 1 4
1  2 3 
a 2
 ax x  .
=    3
a  2
 3 
0
Q.3 If ƒ : R  R, g : R  R are continuous
1 
a
3
a3 
 a
2
functions then the value of the integral
=    0  . / 2
a  2
 3 
 6
e
I I
 [(ƒ( x )  ƒ( x ))(g( x ) – g( x ))]dx
 (log x )  / 2
m
Q.1 Given Im = dx . If m + m 2
1 K L is -
= e then values of K and L are - (A) 1 (B) 0 (C) –1 (D) 
Sol.[B] Let F(x) = [ƒ(x) + ƒ(–x)) (g(x) – g(–x)) so
1 1 F(–x) = (ƒ(–x) + ƒ(x)) (g(–x) – g(x)) = – F(x)
(A) 1 – m , (B) ,m
m 1 m /2

(C)
1
,
m ( m  2)
(D)
Hence  ƒ( x ) dx = 0. (using Property 11).
1 m 1 m  / 2
Q.4 Given f an odd function periodic with period 2
m
,m–2 x
m 1 continuous  x and g(x) =  ƒ( t )dt , then -
0
(A) g(x) is odd function

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 (B) g(2n) = 1
(C) g(2n) = 0  x ( t  1), when x  t
(D) None of these 
x2 t ( x  1), when t  x and t is continuous function
Sol.[C] g(x + 2) =  f (t) dt 1
0 If g(x) =  f ( t ) G(x, t)dt, then which is
2 x 2 x 0

 f ( t ) dt +  f ( t ) dt = g(2) +  f ( t ) dt incorrect
(A) g(0) + g(1) = 0 (B) g(0) = 0
0 2 0
 g(x + 2) = g(2) + g(x)  g(x) is periodic with (C) g(1) = 1 (D) g  (x) = f(x)
1
period 2
2 1 Sol.[C] G(0, t) = 0 for t  0 so g(0) =  f (t) . 0 dt
Also, g(2) =  f ( t ) dt =  f ( t ) dt +
=0
0
0 0
2 G(1, t) = t . (1 – t) = 0 for t < 1.
 f ( t ) dt 1

1 Hence g(1) =  f ( t ) . 0 dt = 0
1 0 0
=  f ( t ) dt +  f (t) x 1

0 –1 Also, g(x) =  f ( t ) t(x – 1) dt +  f ( t ) x (t

dt 0 x
[putting t = u + 2] – 1) dt
1 x 1

=  f ( t ) dt = 0 [f(x) is odd] = (x – 1)  t f ( t ) dt + x  f (t)

–1 x x
 g(2n) = 0 [ g(x) is periodic with period 2] (t – 1) dt
x
Q.5 Let f(x) be positive, continuous and
differentiable on the interval (a, b) and Henc, g(x) = (x – 1) x f(x) +  tf ( t ) dt
0
lim f(x) = 1, lim f(x) = 31/4. If f (x)  f 3(x)
x a  x b – 1

1 +  f ( t ) (t – 1) dt – xf (x) (x – 1)
+ then the greatest value of b – a is - x
f (x ) 1 1

(A) 1 (B) 31/4 (C) (31/4 – 1)(D)

 =  tf ( t ) dt –  f ( t ) dt
4 x x

1 f ( x ) f ( x ) Thus g(x) = f(x).

Sol.[D] f  (x)  f 3 (x) +  4 1 Q.9 For x  R and a continuous function f, let
f (x ) 1  f (x)
1 cos 2 t
Integrating on the interval (a, b), we get
b
I1 =  xf {x(2 – x)} dx and
f ( x )f ( x ) sin 2 t
 1  f 4 (x) dx 
1 cos 2 t
a I1
b I2 =  f {x (2  x )}dx. Then I2
is -
 dx (A) 0
sin 2 t
(B) 1 (C) 2 (D) 3
a
1 b 1 cos 2 t
 b–a
 xf ( x (2  x )) dx
tan–1 f 2 (x )
Sol.[B] I1 =
2 a

sin 2 t
1  lim tan 1 f 2 ( x )  lim tan 1 f 2 ( x )
 b–a   
x a   1 cos 2 t
2 x b 

=  (2  x ) f(x(2 – x)) dx = 2 . I 2 – I1
= . sin 2 t
24
I1
Q.8 Let 2I1 = 2I2  = 1.
G(x, t) = I2

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 1 x2 6 x (1  5x12 ) 6 x (1  5x12 )
lim
Q.10 x 0
x3
 sin t dt is equal to - (A)
(1  x12 ) 2
(B)
(1  x12 ) 2
0
1 2 1 2 6 x (1 – 5x12 )
(A) (B) (C) – (D) – (C) – (D) None of
3 3 3 3 (1  x12 ) 2
x2 these
lim 1
Sol.[B] x 0 3
x
 sin t dt x3
dt
0 Sol.[A] We have, ƒ(x) =  1 t4
x2 1

= xlim  sin t dt  0 
 form 
 1 
 ƒ(x) = 
d
(x 3 ) =
1
0  0  4 .
0
1  t  t  x 3 dx 1  x12
x3 . 3x2
 x 2
  (1  x12 ) . 2 x – x 2 .12x11 
 
  sin t dt  0 at x  0   ƒ(x) = 3  

 0 
  (1  x12 ) 2 
d 6 x (1  x12 – 6 x12 )
[sin t ]t  x 2 . (x 2 ) ƒ(x) =
= xlim
0 dx . (1) (1  x12 ) 2
2
3x 6 x (1 – 5x12 )
(sin x ) ( 2 x ) 2 ƒ(x) =
= xlim = (1  x12 ) 2
0 2 3
3x d  e sin x 
Hence (B) is the correct answer. Q.13 Let F(x) =  ,x>0
dx  x 
x2  
1
 (tan t ) 2 dt 4
3
x
3
0 If e sin x dx = F(k) – F(1), then one of
Q.11 lim is equal to - 1
x 0 x4
the possible values of k is -
 sin t dt
(A) 15 (B) 16 (C) 63 (D) 64
0
d sin x
(A) 1 (B) –1 (C) –1/2 (D) 1/2 e
Sol.[D] F(x) = ,x>0
x2 dx x
1
 (tan t ) 2 dt
0 
On integrating both sides of above equation, we get
lim 0  form  e sin x
Sol.[D]
x 0 x4 0 
F(x) =
 x dx …(1)

 sin t dt 4
3 sin x 3
4
3x 2 3
0 Also,  x
e dx =  x 3
e sin x dx
d 1 1
[(tan 1 t ) 2 ]t  x 2 . (x 2 ) = F (k) – F (1)
= xlim
dx
0 Let x3 = z  3x2 dx = dz
d
[sin t ]t  x 4 . ( x 4 ) 64 sin z
e
[Using L Hospital's Rule]
dx   z
dz = F(k) – F(1)
1
(tan 1 x 2 ) 2 .2 x Using equation (1), we get
= xlim
0
sin x 2 .4 x 3 [F(z) ]164 = F(k) – F(1)
2 F(64) – F(1) = F(k) – F(1)
1 lim  tan 1 x 2   sin x 2  1
  .
= k = 64.
2 x 0  x 2   x2  2
  Hence (D) is the correct answer.
Hence (D) is the correct answer. Q.14 If ƒ(x) is continuous for all real values of x, then
x 3 n 1
Q.12 If ƒ(x) = 1 t4
dt
, then ƒ(x) is equal to -   ƒ (r – 1 + x) dx is equal to -
r 1 0
1

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 n 1  2
n2 1 
lim  1  n
(A)  ƒ( x ) dx (B)  ƒ( x ) dx Q.1 n  3

3
 ...   is
0 0  n (n  1) ( n  2) 8n 
1 1
equal to -
(C) n  ƒ( x ) dx (D) (n – 1)  ƒ( x ) dx
0 0 3 1
n 1 (A) (B)
8 4
Sol.[A]   ƒ (r – 1 + x) dx
r 1 0 1
1 1 1
(C) (D) None of these
8
=  ƒ (x) dx +  ƒ (1 + x) dx +  ƒ (2 + x) 1
0 0 0 n2 n2 1 
dx Sol.[A] nlim

 
3

3
 ...  
1  n (n  1) ( n  2) 8n 
+ …. +  ƒ (n – 1 + x) dx  n2 n2 n2 n2 
0 = nlim

 3
   ...  
1 2 3  (n  0) (n  1) 3 (n  2) 3 (n  n ) 3 
=  ƒ (x) dx +  ƒ (x) dx +  ƒ (x) dx + 1
0 1 2 n n
n2 1
r n = nlim
  (n  r ) 3 = nlim
  n . 1  r 3
…. +  ƒ (x) dx + ….. +  ƒ (x) dx. r 0 r 0
 n
r 1 n 1
n 1 1
dx  1  1 1 
=  ƒ (x) dx. =  3 = 

2
 = –  1 =
2 4 
0 0 (1  x )  2(1  x )  0
Hence (A) is the correct answer.
/2
sin( 2n  1) 3
Q.15 I1(n) =  sin x
dx, I2(n) =
8
.
0
Hence (A) is the correct answer.
/ 2 Q.2 If Sn
sin 2 nx
 sin 2 x
dx, n  N, then-
=
0

(A) I2(n +1) – I2(n) = I1(n) (B) I2(n +1) – I2(n) = I1(n + 1)
 1 1 1 1 
(C) I2(n +1) + I1(n) = I2(n) (D) I2(n +1) + I1(n+1) = I2(n)     ....  
 2n 4n 2  1 4n 2  4 3n 2  2n  1 
Sol.[B] I2(n) – I2(n – 1) =

then nlim S is equal to -

 n
/2 2 2
(sin nx  sin ( n  1)) x
 sin 2 x
dx
   
0 (A) (B) (C) (D)
4 6 3 2
/ 2
sin( 2n  1) x. sin x
=  dx Sol.[B] nlim S =
 n
0 sin 2 x

/ 2
lim
sin ( 2n  1) x n 
=  sin x
dx
0

= I1(n)  1 1 1 1
   ... 
 2 2 2 2 2 2
 4n 0 4n 1 4n 2 4n  (n  1
 I2(n+1) – I2(n) = I1(n+1).
Hence (B) is the correct answer. n 1
1
= nlim


r 0 ( 2n ) 2  r 2

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 n 1
1 Hence (B) is the correct answer.

=
1 lim  2 · /4
r 0  r 
2 n   tan x dx, then nlim
n
1   Q.4 In = 
n [In + In + 2 ]
 2n  0

equals -
1
1
n (A) (B) 1 (C)  (D) zero
2
1
1
1  1  x
1 /4
. 2 sin 1  =  tan
n
= 0 x
2 dx = Sol.[B] In + I n + 2 = x (1  tan 2 x )dx
2 1   2  2 0
0
2
/4

 tan
n
= x sec 2 x dx

. 0
6
1

t
Hence (B) is the correct answer. n
= dt , where t = tan x
100  0
Q.3  (sin 4 x  cos 4 x )dx is equal to -
1
100   In + In + 2 =
n 1
(A) 100  (B) 150 
 lim
(C) 200  (D) None of these n  n [In + In + 2]

100  1 n
= nlim n. = nlim
Sol.[B]  (sin 4 x  cos 4 x )dx = 2  n 1   n 1
100 
n
100  = nlim  1  = 1.
 n 1  
 (sin 4 x  cos 4 x )dx  n 
0
Hence (B) is the correct answer.
(integrand is given)

x 3 cos 4 x sin 2 x

2
( 200) Q.5   2  3x  3x 2
dx is equal to -
=2 0
 (sin x  cos x )dx = 2 . 200
4 4

0 2 2 2 2
(A) (B) (C) (D)
/2 16 8 4 32
 (sin
4
x  cos 4 x ) 
x 3 cos 4 x sin 2 x
0
Sol.[D] Let I =   2  3x  3x 2
dx
dx 0


 4 4  (   x ) 3 cos 4 (  x ) sin 2 (  x )
sin x  cos x is a periodic function of period 2 
 
=   2  3(   x )  3(  x ) 2
dx
0

 / 2 /2
   =
= 400 
 0 
sin 4 x  
cos 4   x  dx 
2  
 0

( 3  3 2 x  3x 2  x 3 ) cos 4 x sin 2 x
/2
3. 1    2  3x  3x 2
d
= 800  sin 4 x dx = 800 .
4 .2
·
2
= 0
0
x
150 .

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  
(  3  3 2 x  3x 2 ) cos 4 x sin 2 x = log 2 – 2 [x – tan–1 x ]10 = log 2 +
=  2
  3x  3x 2
dx – 2
– 2.
0  4  4
y= log 2 –2 = elog 2 . = .
I e 2 e 2 2e 2
/2

( 2  3x  3x 2 ) cos 4 x sin 2 x
 sin
2m
xdx
 2I =   2 2
(  3x  3x )
The value of mlim
0
0 Q.7  /2
2 m 1
dx  sin x dx
/2 0

 cos
4
 I =  . x sin2 x dx = 2 . (A) 0 (B) 1/2
0
(C) 2 (D) None of these
3 .1.1  /2
.
 sin
2n
6 . 4 . 2 2 Sol.[D] We know that I 2n = x dx
0
2
= . 2n  1 2n  3 1 
32 = × × ….. × × ,
2n 2n  2 2 2
Hence (D) is the correct answer. /2
2 n 1 2n 2n  2
1/ n I2n + 1 =  sin x dx = × ×…×
lim 1  1 1  2 .....1  n  2n  1 2n  1
2 2 2
Q.6 0
n  
 n 2  n 2   n 2 
2
is equal to - and
3
 4  4 4
(A) 2 (B) 2 (C) 2m
2e e e 2 Also, I2m + 1 = I2m – 1.
2m  1
 4
(D) 2 2
For all x  (0, /2), sin2m – 1 x > sin2m x > sin2m + 1 x
e
Integrating from 0 to /2, we get I2m – 1  I2m  I2m + 1
Sol.[D]
1/ n I 2 m –1 I 2m
 2  2   2  Whence  1
Let y = nlim  1  1 1  2 .....1  n  I 2 m 1 I 2 m 1
 
 n 2  n 2   n 2 
... (i)
 log
I 2m 1 2m  1
y= Also = .
I 2 m 1 2m
  2   2   2 
lim 1 log1  1   log1  2   ... log 1  n  I 2m 1 2m  1
n  n   2  n 2   n 2  Hence mlim = mlim = 1.
  n      I  2m
2 m 1
n   1
1 r2 From (i) and using sandwitch theorem we have
= nlim
   n2
n r 1
log  1   =
  log (1 +
  0

x2) dx lim I 2m = 1.
m  I 2 m 1
1
2x
= [x log (1 + x2) ]10 –  x . 1 x2 dx
0

1
 1 
= log 2 – 2  1  1  x 2  dx
0

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 1 1 x
1

Q.8 A function ƒ is defined by ƒ(x) = , r < = sin–1 = .
r 1 2 0 6
2 2
b

x
1
, r = 1, 2, 3, …… then the value of
Q.10 Find the value of  sin xdx by first Principle
r 1 a
2
rule -
(A) cos a – cos b (B) sin a – sin b
1 (C) cos b – cos a (D) sin b – sin a
 ƒ( x ) dx is equal - Sol.[A] Here ƒ(x) = sin x & nh = b – a from the
0
definition,
(A) 1/3 (B) 1/4 (C) 2/3 (D) 1/3 b n 1

1  2  ( r 1)
 sin x dx = Lim h
h 0  sin(a  rh)
1 a r 0
Sol.[C]  ƒ( x ) dx =   2 r 1
dx =
= Lim h {sin a + sin (a + h) + sin (a + 2h)
0 r 1 2 r h 0
+ … + … + sin (a + n – 1h)}

1  nh
 2 r 1 [2–(r–1)– 2–r]  sin
2 sin  a  (n  1)h 
r 1 = Lim
h 0 
h
   sin h  2 
=  2  2( r 1) –  2  2r 1  2
1 1
 ba

 2  2r
1 1  sin 2 
= ( 2 2 – 2) =2. .
4 1 1/ 4
=
Lim    sin a  b  a  h 
1 = h 0 h   
 sin h   2 
2  2
.
3 Lim
= h 0
Q.9 The nlim S if
 n h
2.
2 sin  b  a  sin  a  b  a  h 
Sn =
h 2   2
sin   
1 1 1 1 2
   ... 
2n  ba  ba 
4n 2  1 4n 2  4 3n 2  2n  1 = 2 . 1 . sin   sin  
 2   2 
is -
= cos a – cos b.
(A) /2 (B) 2 (C) 1 (D) /6
b

x
2
Sol.[D] nlim S
 n
Q.11 Find the value of dx by first principle rule -
a

= nlim
 1 3 3 1 3 3
(A) (a – b ) (B) (b – a )
3 3
1 3 3
  (C) (a – b ) (D) None of these
  2
1  1 1 1 1 
   ...  
n  40 4 1/ n 2
4  4 / n2
Sol.[B]
2  Here ƒ(x) = x 2, nh = b – a from the definition,
 n 1 
 4  

  n    b n 1

n 1 1  x 2 dx = Lim h
h 0  (a  rh) 2
1 1 dx r 0
= nlim

 = 
a
n r 0 4  (r / n ) 2 0 4  x2

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 n 1 Lim
h 0 
= Lim h (a 2  2ahr  h 2 r 2 ) n 
r 0

= Lim h 
 n n n
h 0    .
2 2

 (3  4 n ) 2 (3 2  4 n ) 3 (3 3  4 n ) 2



n 1 n 1 n 1 
2 = nLim
  a  2ah  r  h  r 
2 2 

r  0 r 0 r 0 

Lim h 
 n n n
= h 0   
 2 2
 1(3 1  4 n ) 2 (3 2  4 n ) 3 (3 3  4 n
n

 2 ( n  1) n h 2 ( n  1) n (2n  1) 
 n
a n  2ah
 2

6


= nLim
  r (3 r  4 n ) 2
  r 1

Lim = Lim
= h 0
h n 

 2 ( nh  h ) ( nh ) ( 2nh  h ) 
a (nh )  a (nh )(nh  h )   n
1
 6    2
r 1  r   r  
= Lim n   3    4 
h 0  n   n  
 

1
 2 ( b  a  h ) (b  a ) ( 2(b  a )  h ) = dx
a (b  a )  a (b  a )(b  a  h ) 
6
  x (3 x  4) 2
  0

(b  a )3 Put 3 x +4=t
= a2 (b – a) + a (b – a)2 +
3 3 dx 2
 dx = dt  dt
(b  a ) 2 2 x x 3
= {b + ab + a2}
3 when x = 0  t = 4
1 3 3 x=1t=7
= (b – a ).
3 7 7
2 dt 2  1
Q.12 Find the limit, when n   of P=
3  t2 =  
3  t 4
4
n n
+
(3  4 n ) 2 2 (3 2  4 n ) 2 2 1 1 
=–   
n 1 3 7 4
+ 2
+ ... +
3 (3 3  4 n ) 49n 2 1 1  2 3
=    = . =
1 1 3 4 7  3 38
(A) (B)
12 13
1
1
(C) (D) None of these 14
14
1
Sol.[C] Let P = Hence P = .
14

( n!)1 / n
Q.13 Find the limit, when n   of
n
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 1 1 (8.6.4.2) (3.1)
(A) e (B) 2
(C) (D) e2 = . 1 (from walli's
e e (13.11 .9.7.5.3.1)

(n!)1 / n formula)
Sol.[C] Let P = nLim

n 128
= .
1/ n 15015
Lim  n !  Lim
n  
=  n = n  n n
n  C
Q.15 Show that : nLim
  n r (r r 3) =e–2
r 0
1/ n
 1. 2 . 3 ... n  (A) e – 1 (B) 2 – e

 n . n . n ... n 

 
(C) e – 2 (D) None of these
1/ n
 1  2  3   n  1
= nLim 
       ...  
 r2 1
Sol.[C]  x dx =
 n  n  n   n  r 3
0
1/ n
 n  r  n 1
= nLim
   
  1
 
 r 1  n  
 nLim
  n
Cr .
n r . x
r2
dx
r 0 0
Taking logarithm then 1
x 
n r
   x . Lim  C r .   dx
2 n
n
1 r 
ln P = nLim
 n
 ln n  0
n 
r 0 n 
r 1
1 n
  x 
  x . Lim 1 
1 2
  dx
=  lnx dx 0

n   n 

0
1
  x e dx
1 2 x

= (1n x . x – x) 0
0
= {x2 (ex) – (2x) (ex) + 2ex} 10 (By
= –1–0
= –1 successive method)

Hence P = e–1 = 1/e. = (e  2e  2e)  (0  0  2)

/2 = e – 2.
Q.14 Evaluate :  sin 9 x cos4 x dx -
0  –1
e tan x
(A)
128
(B)
Q.52 0
1 x2
dx equals -
1515
(A) 1 (B) e/2 +1
(C) e/2 – 1 (D) None of these
128
Sol.[C] Do your self.
15015
64 1
(C) (D) None of

3 x
15015 Q.53 The value of ( x  3e  4) (x2 + ex ) dx is-
0
these (A) (3e + 2)/6 (B) (3e – 2) /6
/2 (C) (3e – 2)2/36 (D) None of these
 sin
9 Sol.[D] Do your self.
Sol.[B] Let I = x cos4x dx
0

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 / 4 =2– 2
 sec
7
Q.54  sin3d=
0
4
(A) 1/12 (B) 3/12
(C) 5/12 (D) None of these Q.68 {
0
x }dx =
Sol.[C] Do your self.
2 7 17
e (A) 1 (B) (C) (D)
Q.55  1/ e
| log x | dx = 3 3 3

 
1 4
(A) e–1 – 1 (B) 2 (1–1/e) 2 2
(C) 1 – 1/e (D) None of these Sol.[C] 
0
x dx +
1
x – 1 dx =
3
+
3
(8 –
Sol.[B] Do your self.
1) – 3
7
=
1
3
x
17 1
Q.56 cos 4 x dx is equal to -
e
x2
Q.69 For f(x) = dx
1
0
(A) –2 (B) 2 1

e
x2
(C) 0 (D) 1 (A) 0  dx  1 (B) 1 
0
Sol.[C] Do your self.
1

e
x2
dx  e 2
/ 2
( x ) 0
Q.57  ( x )  (  / 2 – x )
dx is equal to- 1 1

 e
2
x2
0 (C) 1  e x dx  e (D) 0  dx  e 2
0 0
(A) /4 (B) /2
Sol.[C] For 0  x  1  e  e 0 x2 e
1
(C)  (D) None of these
1
Sol.57 [A] Do your self.
e
x2
 e0 (1 – 0)  dx  e1(1 – 0)
0
Q.58 1

e
2
x
1 dx  e
 1 2
2 2
r 1 2 0
lim   3  ...  3  ....   e2 2
n  13 n 3
2 n 3
r n 3
2n  dx ex
 Q.70 If I1 =  log x
& I2 = 
1
x
dx then -
equals- e

(A) I1 = I2 (B) 2I1 = I2

(A) (1/2) log 3 (B) (1/3) log 2
(C) I1 = 2I2 (D) None of these
(C) 3 log 2 (D) (1/2) log 2
Sol.[A] Put log x = t  x = et

Sol.[B] Do your self. 2 2

et ex
1.5
 I1 = 
1
t
dt (
P –1)
 
1
x
dx = I2

Q.67 
0
[ x 2 ] dx =
a
g( x )
(A) 2 (B) 2 + 2 Q.71 The integral  f ( x )  f (a – x ) vanishes if -
(C) 2 – 2 (D) 0 0

1 2 1.5 (A) g(x) is odd (B) f(x) = f (a – x)

Sol.[C]  [ x 2 ] dx +
 [ x 2 ] dx+  [ x 2 ] dx (C) g(x) = – g(a – x) (D) f(a – x) = g(x)
0 1 2
1 2 1 .5

=  0 dx + 1.dx +  2.dx
0 1 2

=( 2 –1) + 2 (1.5 – 2 )

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 a
g( x ) 1 x m 1
Sol.[C] I = 
0
f ( x )  f (a – x )
then I = .
x m 1
dx
1
n
x
m
=0– .(log x ) n –1 dx
a m 1 0
g (a – x )
 0
f ( x )  f (a – x ) =–
n
Im,n–1
m 1
2
a
g ( x )  g (a – x ) x cos x ex
 2I = 
0
f ( x )  f (a – x ) Q.62 If f(x) = sin x x2 sec x
tan x 1 2

a  / 4
1 g ( x )  g (a – x )
I=
2 
0
f ( x )  f (a – x ) Then  f ( x )dx =
–/ 4

(A) 0 (B) 1
= 0 (vanishes) if –g(x) = g(a – x)
(C) 4 (D) None of these
Sol.[A] f(x) is odd function
Q.72 The value of
 / 4
 
 f ( x )dx
2 –2
  1 x  1– x    = 0.
 pn 

– 2
 1 – x
  qn 
  1  x
  r  dx
 

–/ 4

/2
depends on - cos 
(A) p (B) q Q.41 
0 4 – sin 2 
d is equal to -
(C) r (D) q & p
2 –2   
 1 x  1– x   (A) (B)
Sol.[C] 

– 2
pn 
 1 – x


 qn 
 1  x
    dx
 

2 6
 
2 2 (C) (D)
1 x  1– x  3 5
= p n 
1– x 
 dx–2q n   dx +
 1 x  
–2 –2 Sol.[B]sin  = t cos  d = dt
1

2 
0
dt / 22 – t 2 = [sin–1(t/2)]01  /6

–2
  dx
Q.42 Suppose f is such that f(– x) = – f(x) for every
= p × 0 – 2q × 0 + 4 = 4 1
real x and 0
f ( x )dx  5 , then
0

1  –1
f ( t )dt is equal to -

x
m n
Q.73 Im,n = (nx ) dx (A) 10 (B) 5
0 (C) 0 (D) – 5
n n 0
(A) –
m
Im,n–1 (B) –
m 1
Im,n–1 42.[D]  –1
f ( t )dt put t=–udt=–du–
n 1
(C) – Im–1,n–1 (D) None of these 0
m  1
f (– u )du
1

x
m
Sol.[B] Im,n = (log x ) n dx
0
1 1
 x m1 
 n (log x )
n –1
= (log x ) n .  – .
 m  1  0
0

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 1
 0 f (– u )du   f (– x )dx [prop.
1

0 (A) I1 (B) I1
2
b b
2
 a
f ( x )dx =  a
f ( u )du ]  (C)

I1 (D) 2I1

1 –a

 (– f ( x )dx [f(x) is odd] = – 5 Sol.[C]I1= a xf (sin x ) dx A+B–x=a+(–a)–

0

x=–x
/ 2 2 sin x
Q.43 The value of
 0 2 sin x  2 cos x
dx is - f(–x)=f[sin(–x)]=f(sin x)I1=–a+a/2

–a
(A) 2 (B)  a
f (sin x )dx I1 = /2 I2  I1/I2 = /2

(C) /4 (D) 2 x2

Sol.[C]I = 
/ 2
2 sin x / 2 sin x  2 cos x dx ...(1) Q.47 lim
x 0
 0
tan t dt
is equal to -
0 3
x
/2
By prop. I= 0 2 cos x
/2 cos x
2 sin x
dx..(2) (1)&(2)I= 2 3
(A) (B)
3 2
/4
(C) 1 (D) None of these
2

Sol.[D] xlim
x

Q.44

 
a 
xdx   (a + 4), then - 0  0
tan t dt / x 3 0/0 form
 0 

(A) 0  a  4 (B) – 2  a  4 App.L-H xlim

 0 tan x 2 .2 x – 0 / 3 x 2

(C) – 2  a 0 (D) a  – 2 or a  4

2/3 xlim
0 tan |x|/x
2
Sol.[B] ( x / 2) a0  a + 4  a2  2a + 8 a2–2a–8 0
R.H.L. 2/3 xlim
0
tanx/x= 2/3 limit does not exist.
 (a – 4) (a + 2)  0  – 2  a  4

L.H.L 2/3 xlim

0
tan(–x)/x= – 2/3
1 1
Q.45 The value of I=  0
xx–
2
dx is -
1000
Q.48  0
e x –[ x ] dx is -
(A) 1/3 (B) 1/4
(C) 1/8 (D) none of these e1000 – 1
(A) e1000 – 1 (B)
1/ 2 e –1
Sol.[C]  0
– [ x ( x – 1 / 2)] dx +
e –1
(C) 1000 (e – 1) (D)
1 1000
 1/ 2
x ( x – 1 / 2) dx
1 1
Sol.[C]1000  e x –[ x ] dx 1000  e x dx =1000
0 0
– [ x 3 / 3 – x 2 / 4]10/ 2 + [ x 3 / 3 – x 2 / 4]11/ 2 = (e – 1)
x
1/8 Q.49 Let f(x) = 
1
2 – t 2 dt . Then real roots of
2
the equation x – f '(x) = 0 are -
–a 1
Q.46 Let I1 =  a
xf (sin x )dx , I2 = (A) ± 1 (B) ±
2
–a
1
 a
f (sin x )dx , then I2 is equal to - (C) ± (D) 0 and 1
2
Sol.[A]f'(x)= 2 – x 2 x2=F(x)x2= 2 – x 2 x4=2x2
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 x4+x2–2=0(x2+2)(x2–1)=0x2+20x= ± 1 / 2 / 2
 0
sin xdx   0
sin 3 xdx  I1 > I2
n
lim 1 r
Q.50 n 
n
 r 1 n  r2
2
is equal to - Q.54 If f(x) =
sin x  sin 2 x  sin 3x sin 2 x sin 3x
(A) 1 – 2 (B) 2 – 3  4 sin x 3 4 sin x
1  sin x sin x 1
1
/ 2
(C) 2 (D) – 2 , then the value of  0
f ( x )dx is -

n 2
Sol.[B] nlim

1/n r / n
r 1
1  ( r / n ) 2  (A) 3 (B)
3
1
(C) (D) 0
1 3
0
x/ 1  x 2 dx 1 + x = t  xdx=tdt 
2 2

Sol.[C]C2  C2 + C3
2
f(x) =
1
tdt / t = 2 –1

2 sin x  sin 2 x  sin 3x sin 2 x  sin 3x sin 3x

Q.51  –2
sin 3 x | x | dx is equal to - 3  4 sin x 3  4 sin x 4 sin x
1  sin x sin x  1 1
(A) 0 (B) 1
(C) 4 (D) None of these C1  C1 – C2

Sol.[A]f(x) = sin3x|x| f(–x) = sin3(– x) |– x| sin x sin 2 x  sin 3x sin 3x

f(x) = 0 3  4 sin x 4 sin x
= – sin3x|x| f(– x) = – f(x)  odd fn  = 0
0 1  sin x 1

f(x) = sin x (3 – 4 sin2x) = 3sin x – 4sin3x f(x)

1
Q.52  ({x}[ x  1])dx
0
(where {.} denotes = sin 3x
fractional part of x and [.] denotes greatest /2
  sin 3xdx = – [cos 3x / 3]0 / 2 = 1/3
0
integer of x) is equal
x
to - Q.55 If f(t) is an odd function, then 0
f ( t ) dt is
(A) 1 (B) 0
-
1
(C) (D) None of these (A) an odd function
2
(B) an even function
1 1
Sol.[C]  0
( x – [ x ])([ x ]  1)   0
( x )(1)dx (C) neither even nor odd
(D) 0
0 < x < 1  [x] = 0  ( x 2 / 2)10 = 1/2 x
Sol.[C]f(t)  odd fn then  0
f ( t ) dt is even fn

Q.53 I1 =  0
/ 2
sin xdx , I2 = 0
/2
sin 3 xdx , Q.89  
sin mx sin nx dx , (m  n and m, n are

then - integers) =

(A) I1 > I2 (B) I1 < I2 

(A) 0 (B)  (C) (D) 2 [A]
(C) I1 = I2 (D) I2 = 0 2

Sol.[A]0 < x < /2  sin x > sin3x 2

Q.90  0
x 2 [ x ]dx =

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 5 7 8 4 1
 1 
(A)
3
(B)
3
(C)
3
(D)
3
[B] Q.120 If 
sin x
t2 (f(t)) dt = (1 – sin x), then f 
 3
 is

(A) 1/3 (B) 1/ 3 (C) 3 (D) 3

Q.107 If f : R  R and g : R  R are continuous, then
[C]
/ 2
  / 2
(f(x) + f(–x)) (g(x) – g(–x)) dx =
 sec 2 x 
(A)  (B) 1 (C) –1 (D) 0 [D] 
 2 ƒ( t )dt 

Q.91 lim   
x  / 4  2  
x – 
 16 
Q.109 Let f(x) be a function satisfying f '(x) = f(x) with 8 2
(A) ƒ( 2) (B)
f(0) = 1 and g(x) be a function that satisfies  
f(x) + g(x) = x2. Then the value of the integral 2
(C) ƒ(1 / 2) (D) 4ƒ(2) [A]
1 
 f ( x )g( x ) dx is
 / 2
0
Q.92 If 0 x ƒ(sin x )dx  A 0 ƒ(sin x )dx
2 2
e 3 e 3 , then A is –
(A) e +  (B) e – 
2 2 2 2 
(A) (B) 
e2 5
4
e2 5
(C) e +  (D) e –  [B] (C) 0 (D) 2 [B]
2 2 2 2
e 39  sin( loge x )
Q.110 If f is a positive function and
Q.93 1 x
dx 

k
(A) 2 (B) 0
I1 =  xf {x (1  x )}dx
1 k (C) – 2 (D) 2 [A]
k

I2 =
1 k
 f {x (1  x )}dx , where 2k – 1 > 0, Q.94 2
4 dx
{( x  2)(4  x )}


I1 (A) /2 (B) 

then is
I2 (C) 0 (D) None [A]
1
(A) 2 (B) k (C) (D) 1 [C]
2 Q.95 The value of the integral
/2 cot x dx
2
0 (cot x )  (tan x )
is –

Q.119  [2 sin x ]dx (where


[ ] represents the (A) /4 (B) /2
(C)  (D) None [A]
greatest integer function) = 1

(A) 
5
(B) –  (C)
5 Q.81 Value of 
0
| 5x – 3 | dx = 0 -
3 3
1 13
(D) –2 [A] (A) – (B)
2 10
1 23
(C) (D) [B]
2 10
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 /2 (A) 2 (B) – 3 (C) 3 (D) 5 [D]
Q.82  sin4 x cos2 x dx –

0  1 dx
  Q.96 The value of  log  x  
 x  1 x 2
is -
(A) (B) 0
2 4 (A) log 2 (B) log 2
  (C) 0 (D) None of these [A]
(C) (D) [D]
8 32

Q.83 The area bounded by y = x 2, the lines y = 1 and e sin x
y = 2 on the left of y-axis -
Q.97 If I = 
 e sin x  e sin x
dx, then I equals to -
4 4
(A) (2 2 – 1) (B) (1 – 2 2 )  
3 3 (A) (B)  (C)  (D) [A]
2 4
2 2
(C) (2 2 – 1) (D) (1 – 2 2 ) [C]
3 3 x
2 Q.98 If h(x) =  sin4 t dt, then h(x + ) =
Q.92 The value of 
1
| [x] – {x}| dx. Where [.] 0

h (x )
denotes greatest integer function and {.} denotes (A) (B h(x).
fractional part of x = h ( )
7 5 1 3 h()
(A) (B) (C) (D) [B]
2 2 2 2 (C) h(x) – h() (D) h(x) + h() [D]

x
Q.93 Which of the following statement is true du 
b a Q.111 If  u
e 1
=
6
, then the value of x is
(I) 
a
f(x) dx = – 
b
f(x)
-
log 2

dx (A) 4 (B) log 8

a (C) log 4 (D) None of these [C]
(II) 
a
f(x) dx = 0 if f(x) is an even

dx
function
na a
Q.112  (1  x
0
a
)(1  x 2 )
=

(III)  f(x) dx = n  f(x) dx ; if f(a

(A)

(B)

(C)

(D)

0 0 2 4 8 32
+ x) = f(x) [B]
(IV) If f(x)  (x) and both function are
b b / 2
sin 2002 x
integrable in (a, b) then 
a
f(x) dx  
a
g(x)dx Q.113  ( 2003 x  1)(sin 2002 x  cos 2002 x )
dx
 / 2
(A) I, II, IV (B) I, III -
(C) I, III, IV (D) II, III, IV [C]    
(A) (B) (C) (D) [A]
2 4 8 32
/ 4


x y2
Q.94 If f(n) = tann xdx then f(3) + f(1) is -
 
2
Q.114 If 3  cos t dt + sin t dt = 0 then
0
/3 0
1 3
(A) 1 (B) (C) 0 (D) [B] dy
2 2 =
dx
4 4 1
(A) (B)
Q.95 If 
1
f(x) dx = 4, 
2
(3 – f(x)) dx = 7 then 3  cos 2 x
2 y sin( y 2 )

2 3  cos 2 x  3  cos 2 x
(C) (D) [D]

1
f(x) dx = 2 sin( y 2 ) 2 y sin( y 2 )

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 k 0

Q.115 If I1 = 
3 k
xf {x (3  x )}dx and Sol.[C] I =  [x  [x  [x ]]]dx +
–1
k
I1 1
I2 = 
3 k
f {x (3  x )}dx then
I2
=
 [ x  [ x  [ x ]]]dx
0
2 3 1 0
(A) (B) (C) (D) 2 [B]
3 2 2 let I1 =  [x  [x  [x ]]]dx
–1
Q.38 The value of –1  x < 0  [x] = – 1
 [x + [x]] = [x – 1] = [x] – 1 = – 2
3   x   x 2  1 
  tan –1  2
–1 
  x  1
  tan –1 
  x
dx


 [x + [x + [x]]] = [x – 2] = [x] – 2 = – 1 – 2 = – 3
0

is -
 I1 =  – 3.dx
–1
= – 3  x  0–1 = – 3

(A) 2 (B)  Similarly I2 = 0  I1 + I2 = – 3

(C) /2 (D) /4 [A]
1 1
2
x3
 2 dx , I2 = 2
x
Q.96 Let I1 = dx . Then
/3 dx

0 0
Q.39 is equal to -
/6 1 tan x (A) I1 = I2 (B) I1 < I2
(C) I1 > I2 (D) None of these
 
(A) (B) Sol. C] As 0 < x < 1  x2 > x3  2 x > 2 x
2 3

12 2
1 1
2  2 3

 2 dx >  2 dx  I1 > I2
x x
(C) (D) [A] 
3 4 0 0
1
y2
Q.48 For any integer n, the integral Q.97 Let I = e dy . Then
0


2
e cos x
cos3 (2n + 1) x dx has the value - (A) 3 < I < e (B) 2 < I < e
0
(C) 3< I < e/3 (D) 1 < I < e
(A)  (B) 1 2
Sol.[D] 0 < y < 10 < y2 < 1  e0 < e y < e
(C) 0 (D) None of these [C]
1 1

1. dy <
y2
e
2
 1 < ey < e  dy <
 0 0
Q.51 If f() = 2 and  [f ( x )  f ' ' (x )]
0
sin x dx =
1

5, then f(0) is equal to, (it is given that f(x) is  e. dy

0
continuous in [0, ] : 1
y2
(A) 7 (B) 3 1< e dy < e
0
(C) 5 (D) 1 [B]
3

Q.52 
1

–1
[ x  [ x  [ x ]]]dx , where [•] denote Q.116 The value of  [ x ] dx is
1
greatest integer function, is equal to - (A) 1 (B) 2
(A) – 2 (B) – 1 (C) – 3 (D) – 4 (C) 3 (D) None of these
3 2 3
Sol. [C] I =  [ x ] dx = 1. dx +  2 dx
1 1 2
2 3
= x 1
+ 2x 2
=1+2=3
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 /a

 sec
2
1  x (x9 – 3x5 + 7x3 – x) +
1
Q.16 
–1
1  x2
= – / a

/a

 sec
2
   x dx
(A)  (B) (C) – (D) – / a
2 2 3
/a

 sec x dx  2[tan x ]0 / 4
2
Sol. [B] [tan –1
x ]1–1 = –1
tan (1)–tan (–1)=–1
– odd + = 2
4 – / a

     
–  =  tan 4 – tan 0 = 2
 4 2  
/ 4
Q.17 Value of  log (1 + tanx) dx = /3 1
0 Q. 9 The value of the integral  /6 1 tan x
2 
(A)  (B) (C) log2 (D) dx is
4 8
 
log2 (A) (B)
3 6
/ 4

Sol. [D] I =  1og (1
0
+ tanx) = (C)
12
(D) 0
/3
/4 dx /3





 4

log 1  tan  – x  


 / 6 1
sin x 
cos x
cos x  sin x
0 Sol. [C] = /6
/ 4 cos x
 1 – tan x  dx
=  1og
0
1   dx
 1  tan x   
– 
/ 4 / 4 3 6
2 = = 12
  1og 2
2
= 1og = –
(1  tan x )
0 0
2
/ 4 Q.10 The value of the integral  x [x] dx

0
log (1  tan x )
0
([.]G.I.F) is
/4 7 3
(A) (B)
2I=  log 2 .dx =
0
log 2[ x ]0 / 4 I= 2 2
5
 (C) (D) none of these
log2 2
8 2 1 2

Q.19 If f(x) =
9 5 3
x – 3x  7 x – x  1
then
Sol. [B] 0
x[ x ] dx = 
0
x (0) dx   x.1
1
dx
cos 2 x
2
/ 4  x2 
=  
– / 4
 f ( x ) dx equal to -
 2

1

(A) 0 (B) /2

Q.11 The value of the integral
(C) 2 (D) 1
log 3

Sol. [C] Hence f(x) =

x 9 – 3x 5  7 x 3 – 7 x
+ sec2x
 log1 / 3
log ( x  1  x 2 ) dx is
cos 2 x (A) log 3 (B) 2 log 3
(C) 0 (D) none of these

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 log 3 =
Sol. [C] I = 
– log 3
log ( x  1 x 2 )
dx. 
1

0
( x 3 – 3x 2  2 x ) dx –  1
2
( x 3 – 3x 2  2 x ) dx

 x  1 x 2 
f(x) = log  
  1 2
  x4   x4 
log  – x  1  x2 
  – x3  x 2  – x3  x2 
f(–x) =  =  4 
f (x) + f (–x)= log 0 –
  4 1
 
1 x 2   1 x 2  x
 x   – x   2 x 
 
= log {1 + x2 – x2} = log 1 I = 0

Q. 15
 0
e 2 sin    dx =
2 4

/ 4
(A) 2 (B) e (C) 0 (D) 2 2

n
Q.12 If Un = tan x dx , then Un + Un–2 = 2  1 x 1 x
 ex / 2 
0
sin  cos 
1 1 Sol.[C] 0  2 2 2 2  dx
(A) (B) 
n –1 n 1 2
e
t
(sin t  cos t )
2 dt t = x/2 dx
0
1 = 2 dt
(C) (D)
2n – 1
= 2 e t
sin t  
0 = 2 ×0 =0
1
2n  1
/ 4 / 2 2 sin x
Sol. [A] Un =  tan n x dx Q.16  0 2 sin x  2 cos x
dx =
0
/ 4
=  tan n – 2 x (sec2 x –1) dx  
0 (A) 2 (B)  (C) (D)
/ 4 4 2
 0
(tan x ) n – 2 sec 2 xdx – Un–2
Sol.[C] Direct Result
/ 4
 (tan x ) n –1  1
 
Un + Un–2 =  n – 1  0 = n –1 lim
Q. 29 n 

/ 2 cot x 1/ n
Q.13  0 cot x  tan x
dx =  1  2  3  n 
1   1   1  ....1   
 n  n  n  n 
 
(A) 1 (B) –1 (C) (D) =
2 4
 e 4
Sol.[D] Direct result = (A) (B)
4 4 e
2
(C) (D) none of these
2
e
Q. 14 The value of  0
| x 3 – 3x 2  2 x | dx is
Sol.[B] A =
1 1 1
(A) (B)  1  2  n  n
2 4 lim 1   1   ..... 1   
n   n   n   n 
1
(C) (D) none of these log A =
16
Sol.[A] x3 – 3x2 + 2x = 0
x (x2 – 3x + 2) = 0 x (x – 2) (x – 1) = 0 1  1  2  n  
x = 0, 1, 2.
lim  log 1   1   .....1  
n   n  n  n  n 
– + 
– +
0 1 2 1  1  2  n
2 lim  log 1    log1    .....  log 1 
n  n  n  n  n
| x
3
– 3x 2  2 x | dx
n
1  r 
0
lim
n 
 n log 1  n 
r 1

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 1 Let sinx – cosx = t (cosx + sinx) dx = dt
 log (1  x ) dx  x log (1  x )10 0
dt
=
0
1
= –
=
2  1– t 2 =
2 (sin
–1
t ) 0–1
1 –1
 x  1 x dx
0 

2
1 Q. 32 For any integer n, e sin x
cos 3 ( 2 n  1)
x 1 – 1

0
dx
= log2 – x 1 = log 2 – dx =
0
[ x – log ( x  1)]10 (A) –1 (B) 0 (C) 1 (D) 
= log 2 – [1 – log 2] = 2 log 2 – 1 

e
sin 2 x
log A = log 4 – 1 cos 3 ( 2n  1) x dx
Sol.[B] I =
4 0
log A = log (4/e) A = 
e
e
sin 2 x (  – x )
= cos3 {(2n + 1) ( – x)} dx
3 / 2 0
Q. 30  [ 2 cos x ] dx ([.] G. I. F) = 

=
/ 2 sin 2 x
e {cos3 (2n 1)}( – x) dx
5 5 0
(A) (B) – 
3 3
e
sin 2 x
cos 3 ( 2n  1)} x dx
(C) – (D) –2 =–
0
3 / 2 2 / 3 =–I 2 I = 0 I = 0
Sol.[B]
/2
 [ 2 cos x ] dx
= 
/ 2
(–1) dx 

9

 (–2) dx  Q. 33 Find value of  0

[ x  2] dx ,where [.] is
2 / 3
4 / 3 greatest integer function

+ 

(–2) dx 
+
(A) 31 (B) 22
Y
(C) 23 (D) none
3 / 2

 (–1) dx =
–5
3
3
2
y= x
4 Sol.[A] 1
3
1 X
O 1 4 9
2/3 2
0 /3 /2  4/3 3/2 9 9

–1
–2
0
[ x ] dx +  2 dx
0
1 4 a

=  0
0. dx  1
dx   2dx  18
4
= (4 – 1) + 2 (9 – 4) + 18
/4
3 + 10 + 18 = 31
Q. 31  0
( tan x  cot x ) dx is equal to

2  / 2 x sin 2 x
(A)
2
(B)
2
Q. 34  0 cos 4 x  sin 4 x
dx is equal to

(C)
3
(D) none of these 2 2
(A) (B)
2 2 4
/ 4
sin x  cos x 2
Sol.[A]
0
 sin x cos x
dx
=
2 (C)
8
(D) none of these
/ 4
sin x  cos x
0
sin 2 x
dx Sol.[C] f (x) =
sin 2 x
cos x  sin 4 x
4

/ 4
sin x  cos x
=
2 
0 1 – (sin x – cos x ) 2
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 f (/2 – x) = Q.105 The term independent of 'x' in the expansion of
sin (  – 2 x )
cos 4 (  / 2 – x )  sin 4 (  / 2 – x )  1 1/ 3 –1 / 5 
8
 x x  will be :
sin 2 x 2 
=
sin 4 x  cos 4 x (A) 5 (B) 6
/2
 sin 2 x
Question = 4 
4 0 cos x  sin x
4
(C) 7 (D) 8
Sol.[C] Let (r + 1) term
divide numberator and deominator by 8 r
cos4 x 1 8 r r
Tr + 1 = 8Cr   .
  / 2 2 tan x sec 2 x 2 x 3 x5
=
4 0 
1  tan 4 x
dx
for independent of 'x'
tan2 x = t 2 tan x sec2x dx = dt
 8r r
 dt    – =0
=
4

0 1 t 2
=
4
[tan –1 t ]0 =
4
×
2
= 3 5
8 r r
2  – – =0
3 3 5
8
r=5
3
1
/2 so term is = 8C5   = 7
2
 sin
2
Q.119 x cos 2 x dx is equal to :
0 4

 {x}
[x]
  Q.138 dx where {.}  fractional function
(A) (B) 1
16 24
 and [.]  greatest integer function :
(C) (D) None of these
8 13 1
/2 (A) (B)
12 2
 sin
2 2
Sol.[A] x cos x dx using gamma function
0 15
(C) (D) None
{1.}{1}   12
= × =
4 2 2 16 2 3

 ( x  1)1  ( x  2)
2
Sol.[A] I = dx + +
/2 1 2

 sin
10
Q.106 x (6x9 – 25x7 + 4x3 – 2x)dx equals
–/ 2
4
to :
 ( x  3)
3
dx
 3
(A)  (B)
4 13
I=
 12
(C) (D) None of these
2
Sol.[D] function is odd function 
2 x (1  sin x )
a Q.145  1  cos 2 x
dx
 f ( x ) dx = 0 

–a (A) 2/4 (B) 2

(C) 0 (D) /2

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 
2x 2 x sin x
Sol.[B] I =  1  cos

2
x

1  cos 2 x
dx
Q.73 f : R  R, g : R  R are continuous functions.
 The value of integral
x sin x
I = 0 + 2.2 
1  cos 2 x
dx / 2


0
 [f ( x )  f (– x )][g( x ) – g(– x )]dx ,
– / 2
x sin x
I=4  1  cos
0
2
x
dx
is -
 (A) – 1 (B) 0
(   x ) sin x
I=4 
0
1  cos 2 x
dx (C) 1 (D) 

 Sol. [B] [f(x) + f(–x)] [g(x) – g(–x)] = odd.

sin x dx
2I = 4
0
1 
 cos 2 x / 2 /2

 sin 2 x dx and B =  cos

2
1 Q.74 If A = x dx
dt
I = 2
1

1 t2
0 0

and
I = 2  tan 
1 1
t
1
/ 2
 
 sin
3
I = 2    C= x dx , then -
4 4 0

 (A) A + B = 0 (B) A – B = 0
I = 2 × = 2
2 (C) A + C = 0 (D) A – C = 0
a
/ 2
Q.71 If a is such that  x dx  a + 4, then -
0
Sol. [B] A – B =  (sin
2
x – cos 2 x )dx
0

(A) – 2  a  0 (B) – 2  a  4 1  1 
= . – . [walli's formula]
(C) 0  a  4 (D) a  – 2 or a  4 2 2 2 2

a2
Sol.[B]  a +2  a2 – 2a – 8  8 1 1
2
 (a – 4) (a + 2)  0  – 2  a  4
Q.75 If 
0
f ( x ) dx = 1,  xf ( x ) dx
0
= a and

x
x 2
f ( x ) dx = a2 then
 cos
4
Q.72 If g(x) = t dt, then g(x + ) equals -
0
0

1
(A) g(x) + g() (B) g(x) – g()
 (a – x )
2
f ( x ) dx is equal to -
g(x ) 0
(C) g(x) g() (D)
g ( ) (A) 4a2 (B)2a2
x  (C) a2 (D) 0
 cos  cos
4 4
Sol. [A] g(x + ) = tdt = tdt  1 1

 (a – x )  f (x) –
2 2
0 0 Sol.[D] f ( x )dx  a
x 0 0

 cos 4 tdt 1 1

 x.f ( x ) +  x
2
 2a .f ' ( x )
 x 0 0

 cos  cos
4 4
= tdt + tdt = g() + g(x) = a2 .1 – 2a. a + a2 = 0
0 0

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 /2
sin 2 nx 
Q.76 If an = 0
sin x
dx then a2 – a1, a3 – a2, (A)
4
(B) 

(C) 0 (D) 2 [B]

a4–a3,......are in -
(A) AP (B) GP e 39  sin( loge x )
(C) HP (D) None of these Q.93 1 x
dx 
/ 2
 sin 2 ( nx )
Sol. [C] an – an–1 =  
0
sin x
– (A) 2 (B) 0

(C) – 2 (D) 2 [A]

sin 2 ( n – 1) x 
dx
sin x 
 4 dx
1
/ 2
 cos 2( n – 1) x – cos 2nx 
Q.94 2 {( x  2)(4  x )}

=
2  
 sin x
dx

0 (A) /2 (B) 
/ 2
=  sin(2n – 1) xdx 
0
(C) 0 (D) None [A]

/2 Q.95 The value of the integral

 cos(2n – 1) x  1
– = /2 cot x dx
 ( 2n – 1)  0 2n – 1 0 is –
(cot x )  (tan x )
1 1 1
  a2 – a1 = , a3 – a2 = , a4 – a3= ..... are (A) /4 (B) /2
3 5 7
in HP. (C)  (D) None [A]


2 x3
Q.77 The value of  [f {g( x )}]
–1
f {g(x)}.g'(x)dx,
Q. 93  (1  x 2 )9 / 2 dx =
0
1
(A) 2/35 (B) 3/35
where g(1) = g(2) is equal to - (C) 4/35 (D) None of these [A]
x
(A) 0 (B) 1
(C) 2 (D) None of these
Q. 94 Let f(x) =  2  t 2 dt. Then the real roots of
1
g ( 2) the equation x2 – f (x) = 0 are
1
Sol. [A] Let g(x) = t then I =  f (t)
, f (t)dt (A) ±1 (B) ±1/ 2
g (1) (C) ±1/2 (D) 0 and 1 [A]
 [log f ( t )]gg ((12)) =logf(g(2))–log f (g(1)) = 0
/4
[g(1) = g(2)]
 tan
n
Q. 95 If In =  d , then I8 + I6 equals
0
 sec 2 x  (A) 1/4 (B) 1/5

 2 ƒ( t )dt 
 (C) 1/6 (D) 1/7 [D]
Q.91 lim   
x  / 4  2  
x –  
dx
 16  Q.96  [x  ( x 2  1) ]3
=
8 2 0
(A) ƒ( 2) (B) (A) 3/4 (B) 3/2
 
(C) 3/8 (D) None of these [C]
2
(C) ƒ(1 / 2) (D) 4ƒ(2) [A]
 d e sin x
Q.97 Let F(x) = , x > 0. If
dx x
 / 2
Q.92 If 0 x ƒ(sin x )dx  A 0 ƒ(sin x )dx

, then A is –

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 4
2e sin x2 
 x
dx = F(k) – F(1), then one of the (C)
2
(D) [D]
1
x2
possible value of k is-
 cos t
2
(A) 8 (B) 16 dt
Q.102 is equal to
(C) 32 (D) None of these [B] lim 0
x 0 x sin x

 2  1
Q.81   x  dx is equal to where [] is G . I . F, (A) 2
(B) 1
0
e 
(C) 1 (D) Does not exist [C]
(A) 0 (B) 2/e

(C) e2 (D) n 2 [D] dx
Q.103  a
0
2
 x2  7 is equal to
100 
Q.82

0
1  cos 2 x dx  (A)
231  1 
 
2048  a13 
(B)
231  1 
 
2048  a13 
(A) 0 (B) 200
231  1  232  1 
(C) 100 2 (D) 200 2 [D] (C)   (D)  
2047  a13  2047  a13 

2
[B]

 log sin x dx 
10
Q.83 3 x

0
Q.104
10
 3  x
dx is equal to, where [] is G . I . F

  40
(A) log 2 (B)  log 2 (A) 20 (B)
2 2 n 3
(C)  log 2 (D) 0 [B] 20
(C) (D) None of these [B]

n 3
Q.84 e
cos 4 x .
. cos5  2n  1 x dx,  n  I  is n 1

0 Q.105 if  f  x  dx = n2 + n ; n I, then value

equal to n
(A)  (B) 1 of

3
(C)
2
(D) 0 [D]
3
 f ( x )dx is equal to
50 
(A) 6 (B) 10
 sin
3
Q.85 x dx is equal to
(C) 16 (D) 12 [C]
0
n
(A)  (B) 1
 200   [ x ] dx
(C) 50 (D)   Q.106
0
is equal to (where [x] is G . I . F
 3  n

Q.86
[A]  {x} dx
0
1  2 3 n  and {x} is fractional part of x and n N)
lim  sin  sin  sin  ........  sin 
n  n  2n 2n 2n 2n 
n
is equal to (A) 0 (B)
(A) 1/ (B) 2/ 2
(C) 2/ (D) 2 [B]  n 1 
(C)   (D)
 2 
3

Q.101   sin x  dx
0
is equal to, where [] is G .I .F .,  n  1 [D]

3
(A) (B)   x3

2
Q.75 dx equals -
0 (1  x 2 )9 / 2
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 (A) 2/35 (B) 3/35 
1
(C) 4/35 (D) None of these Q.85 
0
2
(1  cos 2 x ) dx is equals to -

Sol.[A] Put x = tan t, then

/ 2 (A) 0 (B) 1
tan 3 t
I = 
0
sec 9 t
sec2 t dt = (C) 2 (D) none of these
 / 2
/ 2

 sin
3
t cos t dt 4 Sol.[C] I = 
0
| cos x | dx =  cos x dx
0
+
0
2 . 3.1 2 
= =
7 . 5 . 3.1 35
/ 2
 (– cos x ) dx
=  sin x  0 –  sin x   / 2 = 1 – ( – 1) = 2
/ 2 
15
dx
Q.76  ( x – 3)
8
x 1
equals
/2
(A) log (5/3) (B) log (3/5)
(C) 1/2 log (3/5) (D) 1/2 log (5/3)
Q.86  | sin x – cos x | dx is equal to
0

Sol.[D] Put x + 1 = t  dx = 2t dt2

(A) 0 (B) 2 –1
4 4
2 t dt dt
I =  (t
3
2
– 4) t
= 2 
3
t2 – 4
= (C) 2( 2 – 1) (D) 2( 2 + 1)
/ 2 / 2

  f ( x ) dx
4
1  t –2 Sol.[C] I = f ( x ) dx +
2 . 2  log 
2  t 23 0 / 4

1 1 1 1 where f(x) = |sin x – cos x|

= log – log  = log(5 / 3)
2  3 5 2 / 4

=  (cos x – sin x ) dx +
0
/ 2
sin x
Q.79 
0
sin x  cos x
dx equals
/ 2

(A) /2 (B) /4 (C)  (D) 2  (sin x – cos x ) dx

/ 4
Sol.[B] Using property P-4, we have
/ 2
=  sin x  cos x  0 +  – cos x – sin x   / 4
/ 4 / 2
cos x
I=
0

cos x  sin x
dx

Adding it to given integral we have  1   1 

/ 2
= 2   – 1 + (– 1) + 2   = 2
 2  2
2I =  dx =
 x  0 / 2
= /2
0 ( 2 – 1)
 I = /4
1 / 4
x sin –1 x
  tan
n
Q.80 dx equals Q.87 If In =  d (n N), then n(In–1 +
2
0 1– x 0

(A) /2 (B) /4 (C) 0 (D) 1 In+1), is equal to -

dx (A) /2 (B)  (C) 1 (D) 2
–1
 dt
Sol.[D] Put sin x = t, then 1– x 2
/ 4

 (tan
/ 2
Sol.[C] In–1 + In+1 =
n –1
  tan n 1 ) d
 t sin t dt  t  – cos t   0 / 2
 I = = + 0
0
 sin x  0 / 2 = 1

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 / 4 (A) 1/3 (B) 1/ 3
 (tan
n –1
=   sec 2 ) d =
0
(C) 3 (D) 3
/4
 tan n   1 Sol.[C] Differentiating both sides with respect to x, we
  = have
 n  0 n 0 – sin2x f(sin x). cos x = – cos x
 n(In–1 + In+1) = 1 1
 f(sin x) =
3 sin 2 x
[cos x  0 as t  (sin x, sin /2) so x  /2]
|1 – x
2
Q.88 | dx is equal to
–2
 f(x) = 1/x2
(A) 7/3 (B) 14/3 (C) 28/3 (D) 1/3  1 
f 
 
 =3
–1 1  3 
Sol.[C] I =  (–1  x 2 ) dx +  (1 – x 2 ) dx + 
log( x  1 / x )
–2 –1 Q.91 I1 = 
0
1 x2
dx and I2 =
3

 (–1  x
2
) dx / 2
1
–1 1
 log sin 2t dt , then I 1 + I2 is equal to -
 x3   x3  0
= – x   + x –  +
 3 
–2 
 3  –1 (A) (/2) log 2 (B) – (/2) log 2
(C)  log 2 (D) –  log 2
3
 x3  / 2
– x  
 3 
1 Sol.[A] I1 =  log(tan t  cot t ) dt ,
0
where x = tan
2 2 2 2  2 t
=    +    +  6   = 28/3 / 2
 3 3   3 3   3
=  log(2 / sin 2t ) dt
0
= (/2) log 2 – I2
 I1 + I2 = (/2) log 2
Q.89

Q.103 If [ . ] denotes greatest integer function, then

 1 1 2 4 1 
lim 2 sec 2 2  2 sec 2 2  ....  sec 2 1 8
n  
[x 2 ]
n n n n n  2
[ x 2 – 20 x  100]  [ x 2 ]
dx is equal to -

equals
(A) 5 (B) 4
(A) (1/2 tan 1) (B) tan 1 (C) 3 (D) none of these
8
(C) 1/2cosec 1 (D) 1/2 sec 1 [x 2 ]
n 2 Sol.[C] I = 
[( x – 10) 2 ]  [ x 2 ]
dx
.....
r 2 r

2
Sol.[A] Limit = nlim

sec (i)
n2 n2 8
r 1 [(10 – x ) 2 ]
n 2
I=   22
dx
.....
  n  sec 2
r r 1 [ x ] [( x – 10 ) ]
= nlim

  . 2
n n (ii)
r 1 8
1
1
  1  dx = 6
 x sec tan x 2 2I= I=3
2
= x 2 dx = 0 = (1/2) tan 2
0
2
1
1
 1 
t
2
Q.90 If f ( t ) dt = (1 – sin x), then f  
sin x  3
is equal to

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 / 2

Q.104 If I1 =  cos(sin x ) dx ,
0
I2 = (A)  (B) a (C)
2
(D) 2

a a

  f ( x )  f ( x )dx
/ 2
Sol.[C] use f ( x ).dx =
 sin(cos x ) dx ,
0
a 0

1/ 2   1  x 
/ 2 Q.110 The integral 1/ 2 [x ]  n 1  x  dx
I3 =  cos x dx , then
0 equals -
(A) I1 > I2 > I3 (B) I3 > I2 > I1 1
(C) I1 > I3 > I2 (D) I3 > I1 > I2 (A) – (B) 0
2
Sol.[C] sin x < x  x > 0
 sin (cos x) < cos x for 0 < x < /2 1
/ 2 /2 (C) 1 (D) 2 n  
2
  sin(cos x ) dx
0
<  cos x dx
0
 I2 < 1/ 2
1 x
1/ 2
I3 .... (i) Sol.[A]  [ x ]  In
1 x
=  [ x ] +0
Further x  [0, /2] sin x < x and x1, x2  [0, /2] 1 / 2 1 / 2
x1 > x2  cos x1 < co s x2 1/ 2
 cos x < cos (sin x) 1
/2 / 2
=  [ x ]  [ x ].dx = (1)10/ 2 = 
2
 cos x dx  cos x (sin x ) dx
0
= <  I3 <
0 0
t2 2 5  4 
I1 ..... (ii)
(i), (ii)  I2 < I3 < I1  I1 > I3 > I2
Q.111 If 0 xf ( x )dx =
5
t , t > 0 then f  
 25 
=
1000 n
2 5 2
(A) (B) (C) – (D) 1
Q.105  e
n 1 n –1
x –[ x ]
dx is equal to 5 2 5
t2
2 5
(A) 1000 (e – 1) (B)
1
(e – 1)
Sol.[A] By Nh.  xf ( x )dx = 5
t
1000 0

1 2 4
(C) (e1000 – 1) (D)  2t.t2.f(t2) = .5t
1000 5
1 2
(e1000 – 1)  f(t2) = t  f(4/25) = as t > 0
e –1 5
1000 n 1 Q.112
Sol.[A] 
n 1 n –1
e x –[ x ] dx
=  e x –[ x ] dx
+
0
 2 ( x
3
 3x 2  3x  3  ( x  1) cos( x  1)dx
0
2

e
x –[ x ]
dx +.....+
is equals to -
1
1000 (A) – 4 (B) 0 (C) 4 (D) 6


999
e x –[ x ] dx
= Sol.[C]
0

 ( x  1)
3
 ( x  1) cos( x  1)
1
2

1000 e x –[ x ] dx
0
let x +1 = t ; dx = dt
1
[ x – [x] is periodic with period = 1]
t
3
1  t. cos dt  0


x 1
e dx
= 1000 = 1000 (e – 1)
0

 cos 2 x
Q.109   1 a x
dx , (a > 0) =

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 10 2 3
Q.113    2n 1 sin 27 x dx + (A) 0 (B)
4
n 1
10 (C) 2 (D) None of these
2 n 1
  2n sin 27 x dx /2
 4  3 sin x 
n 1 Sol.[A] I =  log 
 4  3 sin x 
equals - 0

(A) 27 (B) 54 (C) –54 (D) 0 /2

 4  3 cos x 
2n 
King
 log 
 4  3 sin x 
Sol.[D] let I   sin 27  0

 2 n 1 I=–I I0

2n
I1 =  sin
27
(  t )(dt )  xt
1
dx dt a sin 2 t a 2

2 n 1 Q.118 lim
x 0

x  y
e dt 
xy  e sin t dt  =

2 n 1 2 n 1
(A) 1 (B) 0
 sin  sin
27 27
=   dt =  xdx
2 2
2n 2n (C) e sin y (D) sin 2 y.e sin y

2 n 2 n 1
a a 
 sin 27 x   sin 1 
27
 
2 2
x dx = 0 Sol.[C] h 
x 0 x   e sin t   e sin t dt 

 2 n 1 2n y xy 
xy
2

 e sin t dt
2
( x  y)
2a f (x ) e sin
Q.114 0 f ( x )  f ( 2a  x )
dx - =
h
y
= h
x 0 1
=
x 0 x
(A) a (B) 2a (C) 1 (D) 0
sin 2 y
Sol.[A] e
x 1
Q.119
0 (tan x) 2
Q.115 If f : R → R and g : R → R are continuous, then lim dx =
/2 x  2
x 1
   / 2 (f ( x )  f ( x ))(g ( x )  g( x ))dx
=  2 2
(A) (B) (C) (D) 
(A)  (B) 1 (C) –1 (D) 0 4 2 4
x
Sol.[D] (x) = (f(x)+f(–x) (g(x)–g(–x)) 1
Sol.[C]
0 (tan x ) 2 dx
(–x) = (f(–x)+f(x) (g(–x)–g(x)) h
x 
= –(x) → f(x) = odd fn x2 1
/2 x
d 1
  ( x )  0 dx  (tan x ) 2 dx
 / 2 = 0
h
x  2x

1 1 2 x2 1
x2 x3
Q.116 If I1 = 0 2 dx , I2 = 0 2 dx , I3 =
= (tan 1 x ) 2 x 2  1 
2 x2 2 x3 h
1 1

2 dx and I4 = 2 dx , then - x  x
(A) I3 = I4 (B) I3 > I4 1 2

= h (tan 1  x ) 2 1  =   .1 =
(C) I2 = I1 (D) I1 = I2 x  x2 2
Sol.[D]
2
4
/2  4  3 sin x 
Q.117 0 log
 4  3 cos x


dx =
  x log x
Q.148 Value of 0 (1  x 2 ) 2
dx  -

(A) 1 (B) 0
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 (C) 3 (D) None of these 1
= (b – a)8
Sol.[B] take x = tan 280
/2
  sin . cos . log tan d Q.83 If for a continuous function f
0
a a
n/2
 f ( x )dx  k  (f ( x )  f (  x )) dx,

a 0
I= sin . cos . log cot d
0 then k is
2I = 0  I = 0 (A) 0 (B) 1
(C) 2 (D) –1
2 sin 2 x
Q.149 Value of  2  x  1
dx ,
Sol.[B] use  
a
f ( x ).dx  
a
f (x) + f(–x)dx
  2
a 0

  is true for every f(x)  k = 1

Where [x] is the greatest integer function) is -
(A) 1 (B) 0 Q.84 If
(C) 4 – sin 4 (D) none of these
/ 2 / 2
0
sin 2 xdx
2
sin 2 xdx  cos n x sin n xdx  k  cos n xdx,
 
0 0
Sol.[B] Take I =  =0
1  1/ 2 0  1/ 2 then k is
2 0
(A) 2–n (B) 2–n/2
d e sin x (C) 2–2n (D) none of these
Q.150 Let f (x)  ,x 0
dx x Sol.[A] Do yourself
2
2e sin x
4
If
1 x
dx  f (k )  f (1) , then one
Q.85  0

x log sin x dx =
of the possible value of k is
(A) 4 (B) 16 2
(A) 2 log 2 (B) – log 2
(C) 8 (D) none of these 2
Sol.[B] (C)  log 2 (D) –  log 2
 

5
Sol.[B] Remove ‘x’  I =
Q.81 If  1
f (x) dx = 6.53 – 6(1)3, then f(x) =
2 0
log sin x.dx

3 4 3 2  / 2
(A)
2
x (B)
4
x I=
2
×2 
0
log sin x.dx [P-6]

(C) 18x2 (D) none of these  2

Sol.[C] Do yourself = × [– log 2] = – log 2
2 2

Q.82 Value of 
b
(x  a )3 (b – x)4 dx is
Q.87  8
(sin 193 x  x 295 ) dx is equal to
a
(A) 0 (B) 2(8295 + 1)
1
(A) (b – a)8 (B) (b – a) 8
(C) 8295 + 2 (D) none of these
280
Sol.[A] odd function
1
(C) (b – a)9 (D) none of these
280 x

Sol.[A] use x = a cos  + bsin  2 2 Q.88 The function f(x) =  0

log( t  1 t2 ) dt is

 dx = 2(b – a)sin. cosd (A) an even function (B) an odd function

/ 2
(C) a periodic function (D) none of these

7 
 I = 2(b – a)8 0
sin  . cos  d
Sol] log (t + 1  t 2 ) is odd function
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 x 100
so f(x) =  0
log(t  1 t2 ) . dt
Q.93 
n
e x  [ x ] dx =
n 1
n 1
= even function
e100  1 e 1
(A) (B)
1/ 2 1 x e 1 100
Q.89 If  1 / 2
cos x log
1 x
dx = k log 2, then
e100  1
(C) 100 (e – 1) (D)
100
k is
Sol.[C] x – [x]’s Period = 1
(A) – 1 (B) – 2 100 n 1

(C) 0 (D) none of these


n 1
n 1
e x [ x ] dx  100 0
e x [ x ] dx

1
Sol.[C] odd function = 100 0
e x  0 dx  100(e  1)

 x log x

4
Q.90  2
x 6x dx = ……. Q.94
0 (1  x 2 ) 2
dx =
32 32
(A) (3  2 ) (B) (3  2) (A) 1 (B) 0
5 5 (C) 3 (D) none of these
16 Sol.[B] Put x = tan 
(C) (3  2 ) (D) none of these
5
Sol.[B] Put x  6 – x Q.95 If I1 =
1 1 2
  
x2 x3 x2
2 dx , I 2  2 dx, I 3  2 dx
2 0 0 1
d x
Q.91
dx x  cos t dt = ……..
and I4 =  1
2 3
2 x dx , then
(A) 2x cos x2 – cos x
(A) I3 = I4 (B) I3 > I4
(B) 2x cos x 2  cos x (C) I2 > I1 (D) I1 > I2
(C) 2x cos x2 – cos x Sol.[D] Do yourself
(D) none of these
Sol.[B] use NL Sol.[B] Do yourself
3
Q.92  x sgn (sin x) dx = ……..
cos 2 x
0

where sgn denotes signum of x. Q.97 The value of
  1 a x
dx, a > 0, is
3 2 2
(A)  (B)  (A) /a (B) 2
2 3
(C) a (D) /2
3 2 Sol.[D] Do yourself
(C)  (D) none of these
2 1 dx
Sol.[C] I = Q.98  0 2
x  2 x cos   1
=
 2
 0
x. sgn(sin x )dx  
x. sgn(sin x )dx (A) sin  (B)  sin 
 
(C) (D) sin
3
2 sin  2
  2
x sgn(sin x )dx

= 1 dx
 2 3
Sol.[C] I =  0 ( x  cos ) 2  sin 2 
 0
x.1dx  
x ( 1) dx  2
x.(1)dx
1
1  1  x  cos   
 1 2 1 3 2   tan  
=  (42 – 2) + [92 – 42] = sin    sin   0
2 2 2 2

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 = 1 x  ( x  1)
=  0
tan 1
1  x ( x  1)
dx
1  1  1  cos   1  cos  
 tan    tan   1
sin    sin    sin   =  0
(tan 1 x  tan 1 ( x  1)) dx
1 1
= =  0
tan 1 x  
0
tan 1 (1  x  1)dx

1  [P-4]
   
  cot 1  cot     cot 1 (cot )  1
sin   2  2 2  =2  0
tan 1 x dx

1    
=    
sin   2
2
 2 sin  Q.103  (sin x  | sin x | ) dx =
0

(A) 4 (B) 0
e dt
Q.99 Value of 
1/ e t3  t
is equal to (C) 1
Sol.[A] Do yourself
(D) 8

1
(A) 2
(B) e2 – 1
e 1 /2 dx
(C) – 1 (D) 1
Q.104  0 1  tan 3 x
=
e dt e1 t 
 
(A) 0 (B) 1
Sol.[D] I = 2
   2 dt
1 / e t ( t  1) 1/ e  t t 1  (C) /2 (D) /4
Sol.[D] Do yourself
=  log t  1e/ e 
1
2

log( t 2  1)  e
1/ e 1
5 x2
Q.100 Value of 
b x
dx, a < b < 0 is
Q.105
 3 x2  4
dx =
a |x|
 15 
(A) – | a | – | b | (B) | b | – | a | (A) 2 – loge  
 7 
(C) | a | – | b | (D) | a | + | b |
Sol.[B] Do yourself  15 
(B) 2 + loge  
Q.101 Value of  7 

  (C) 2 + 4 loge 3 – 4 loge 7 + 4 loge 5

2 1 x  1 x 
 p n 
2 
  q n 
1 x 
  r
1  x  
dx
 15 
(D) 2 – tan–1  
depends upon  7 
(A) p (B) q Q.106
(C) r (D) p and q
0
2  1 x  1 x    ( x 3  3x 2  3x  3  ( x  1) cos( x  1))
Sol.[C]  
2 
p n    q n 
 1 x 
  r
 1 x   dx
2

dx is equal to
odd odd
(A) – 4 (B) 0
 Answer depends on ‘r’
(C) 4 (D) 6
1  1 
Q.102  0
tan 1  2  dx is
 x  x 1
Sol] Do yourself

(A) n 2 (B) – n 2 2

 
Q.101 2
| [ x ] | dx is equal to
(C) + n 2 (D) – n 2
2 2 (A) 1 (B) 2
1 1 (C) 3 (D) 4
Sol.[D]  0
tan 1
1  x ( x  1)
.dx Sol.[D]

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 2 1 0 1 2
 2
| [ x ] | dx  2
| [ x ] | dx  1
| [ x ] | dx    | [ x ] | dx 
0

1
| [ x ] | dx

1 0 1 2
=  2
2dx   1
dx   0
0.dx   1
dx

= 2 x   2   x  1  0   x 
1 0 2
1

= 2(–1 + 2) + (0 + 1) + 0 + ( 2 –1)
=2+1+1+=4

Q.66 The value of the integral

1
d  1
 dx  tan
1
 dx is-
1
x
 
(A) (B)
2 4

(C) – (D) None of these
2

Q.66 The value of the integral

1
d  1
 dx  tan
1
 dx is-
1
x
 
(A) (B)
2 4

(C) – (D) None of these
2



Q.

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