Question.1
Let f (x) be a continuous function such that f (a –x) +
f (x) = 0 for all x ∈ [0, a], then
is equal to
(A) a
(B)
a/2
(C)
f(a)
(D) 
Solution
I =
=
(
f (a –x) + f (x) =
0 given)
= 
2I = 
2I = a
&⇒ I = 
Question.2
is equal to
(A) 0
(B) 2
(C) e
(D)
none of these
Solution
I = 
Here
the integrand is odd function. .
hence
I = 0
Question.3
If I =
d x then
(A) 0
(B) 2
(C)
(D) 2
– 
Solution
I
= 
I =
d x = 2 
= 2 –
2 tan–1 x
= 2 –
.
Question.4
Let
f : R → R and g : R → R be continuous functions. Then the
value of .
(f (x) + f (-x)) (g (x) – g (-x)) dx is
(A) p
(B) 1
(C)
–1
(D) 0
Solution
Since h (x)
= ( f (x) + f (-x) (g (x) – g (-x)) is an odd function, the value of the given
integral is zero. .
Question.5
The value of
( where {x} is the
fractional part of x) is
(A) 50
(B) 1
(C) 100
(D) none of these
Solution
Given integral =
( by the def. of {x} )
=
-
=
where t2 = x
=
= 
= 
Question.6
If
I =
, then
(A) 0
< I < 1
(B) I
>
(C) I
<
(D) I
> 2 p
Solution
Since
x ∈ 
&⇒ 1 £ 1 + sin 3 x £ 2
&⇒
£
£ 1
&⇒ 
&⇒ 
Question.7
f (sin x) dx is equal to
(A)
dx
(B)
dx
(C)
dx
(D)
none of these
Solution
I
=
f (sin x) dx =
f sin (p
– x) dx = p
(sin
x) dx – I
&⇒ I = 
f (sin x) dx
Again,
I = 
f (sin x ) dx = 2 
= p 
Question.8
The
value of
|sin 2 p x| dx is equal
to
(A) 0
(B)
(C)
(D) 2
Solution
Since
|sin 2 p x | is periodic with period 1/2,
I =
|sin 2 p x| dx = 2
sin 2 p x dx = 2
=
Question.9
If
f(x) =
, then
is equal to ( [ . ]
denotes the greatest integer function)
(A)
(B)

(C)
-6
(D)

Solution
3 [x] - 5
= 3[x] –5 , if x > 0
= 3[x] + 5, if x < 0
&⇒ 
= -1
=
.
Question.10
is
equal to
(A)
(1) g (1) – f (1)
(1)
(B) f
(1)
(1) +
(1) g (1)
(C) f
(1)
(1) –
(1) g (1)
(D)
none of these
Solution
Integrating
by parts, we get
f (x)
(x)dx –
(x) g(x)dx
= f
(x)
(x) –
(x) .
(x) dx –
(x) g (x) + 
(x)
(x) dx
= f
(x).
(x) –
(x) g (x)
Hence
f (x)
(x)
dx – 
(x) g (x) dx = f (1)
(1) –
(1)
g (1)