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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 7
∫
π
/
4
0
log
(
1
+
tan
2
θ
+
2
tan
θ
)
d
θ
=
Report Question
0%
π
log
2
0%
(
π
log
2
)
/
2
0%
(
π
log
2
)
/
4
0%
log
2
If
∫
2
x
√
1
−
4
x
d
x
=
K
sin
−
1
(
2
x
)
+
C
,
then K is equal to
Report Question
0%
ℓ
n
2
0%
1
2
ℓ
n
2
0%
1
2
0%
1
ℓ
n
2
Explanation
I
=
∫
2
x
√
1
−
4
x
d
x
Let
2
x
=
z
⇒
2
x
log
2
d
x
=
d
z
⇒
I
=
∫
1
log
2
√
1
−
z
2
d
z
=
1
log
2
⋅
sin
−
1
z
+
c
=
1
log
2
⋅
sin
−
1
(
e
x
)
+
c
=
k
sin
−
1
(
2
x
)
+
c
then
k
=
1
log
2
.
The value of the defined integral
∫
π
/
2
0
(
sin
x
+
cos
x
)
√
e
x
sin
x
d
x
equals
Report Question
0%
2
√
e
π
/
2
0%
√
e
π
/
2
0%
2
√
e
π
/
2
.
cos
1
0%
1
2
e
π
/
4
Evaluate
∫
π
/
2
0
cos
x
(
1
+
sin
2
x
)
d
x
Report Question
0%
π
2
0%
π
4
0%
π
0%
None of these
Explanation
Let
I
=
∫
π
/
2
0
cos
x
(
1
+
sin
2
x
)
d
x
Put
sin
x
=
t
and
cos
x
d
x
=
d
t
.
[
x
=
0
⇒
t
=
0
]
and
[
x
=
π
2
⇒
t
=
1
]
.
∴
I
=
∫
1
0
d
t
(
1
+
t
2
)
=
[
tan
−
1
t
]
1
0
.......
∫
d
x
x
2
+
a
2
=
tan
−
1
x
a
+
C
=
π
4
−
0
.
=
π
4
.
Evaluate
∫
√
8
√
3
x
√
1
+
x
2
d
x
Report Question
0%
19
3
0%
19
6
0%
38
3
0%
9
4
Evaluate
∫
π
/
2
0
cos
3
x
d
x
Report Question
0%
1
0%
3
4
0%
2
3
0%
None of these
Evaluate :
∫
1
0
√
1
−
x
1
+
x
d
x
Report Question
0%
π
2
0%
(
π
2
−
1
)
0%
(
π
2
+
1
)
0%
None of these
Evaluate:
∫
1
1
/
3
(
x
−
x
3
)
1
/
3
x
4
d
x
=
Report Question
0%
3
0%
0
0%
6
0%
4
Explanation
I
=
∫
1
1
/
3
x
1
/
3
(
1
−
x
2
)
1
/
3
x
4
d
x
Let
x
=
sin
θ
,
d
x
=
cos
θ
d
θ
I
=
∫
π
/
2
sin
−
1
1
/
3
(
sin
θ
)
1
/
3
(
cos
2
/
3
θ
)
sin
4
θ
×
cos
θ
d
θ
=
∫
cos
5
/
3
θ
sin
11
/
3
θ
d
θ
×
cos
2
θ
cos
2
θ
=
∫
cos
11
/
3
θ
sin
11
/
3
θ
cos
2
θ
d
θ
=
cot
11
/
3
θ
.
sec
2
θ
d
θ
Let
tan
θ
=
t
⇒
sec
2
θ
d
θ
=
d
t
=
∫
t
−
11
/
3
d
t
=
t
−
8
/
3
(
−
8
/
3
)
=
−
3
8
t
−
8
/
3
=
−
3
8
(
tan
θ
)
−
8
/
3
|
π
/
2
sin
−
11
1
/
3
As
sin
−
11
/
3
=
tan
−
1
1
2
√
2
⇒
I
=
−
3
8
(
1
(
∞
)
8
/
3
−
(
1
2
√
2
)
−
8
/
3
)
=
3
8
×
(
2
√
2
)
8
/
3
=
3
8
(
2
3
/
2
)
8
/
3
=
3
8
×
16
=
6
⇒
(
C
)
.
Evaluate
∫
a
0
x
d
x
√
a
2
+
x
2
Report Question
0%
a
(
√
2
−
1
)
0%
a
(
1
−
√
2
)
0%
a
(
1
+
√
2
)
0%
2
a
√
3
Explanation
∫
a
0
x
d
x
√
a
2
+
x
2
Let
t
=
(
a
2
+
x
2
)
⇒
d
t
=
2
x
d
x
when
x
=
0
,
t
=
a
2
when
x
=
a
,
t
=
2
a
2
=
1
2
∫
2
a
2
a
2
2
x
d
x
√
a
2
+
x
2
=
1
2
∫
2
a
2
a
2
d
t
√
t
=
1
2
∫
2
a
2
a
2
t
−
1
/
2
d
t
=
1
2
[
t
−
1
/
2
+
1
−
1
2
+
1
]
2
a
2
a
2
=
1
2
[
√
t
1
2
]
2
a
2
a
2
=
2
2
[
√
2
a
2
−
√
a
2
]
=
a
(
√
2
−
1
)
The correct evaluation of
∫
π
/
2
0
sin
x
sin
2
x
is
Report Question
0%
4
3
0%
1
3
0%
3
4
0%
2
3
Explanation
I
=
∫
π
/
2
0
sin
x
sin
2
x
d
x
I
=
∫
π
/
2
0
2
sin
2
x
cos
x
d
x
u
=
sin
x
⇒
d
u
=
cos
x
d
x
=
∫
1
0
2
u
2
d
u
=
2
u
3
3
|
1
0
=
2
3
∫
e
x
3
+
x
2
−
1
(
3
x
4
+
2
x
3
+
2
x
2
x
=
h
(
x
)
+
c
then the value of
h
(
1
)
h
(
−
1
)
.
Report Question
0%
1
0%
-1
0%
2
0%
-2
∫
π
0
x
2
(
1
+
s
i
n
x
)
2
d
x
equals
Report Question
0%
π
(
π
−
2
)
0%
π
2
(
π
−
2
)
0%
π
(
2
−
π
)
0%
none of these
The value of the integral
∫
π
/
2
−
π
/
2
(
x
2
+
log
π
−
x
π
+
x
)
cos
x
d
x
is
Report Question
0%
0
0%
π
2
2
−
4
0%
π
2
2
+
4
0%
π
2
2
Explanation
I
=
∫
π
/
2
−
π
/
2
[
x
2
+
l
o
g
(
π
−
x
π
+
x
)
]
c
o
s
x
d
x
As,
∫
a
−
a
f
(
x
)
d
x
=
0
,
when
f
(
−
x
)
=
−
f
(
x
)
∴
I
=
∫
π
/
2
−
π
/
2
x
2
c
o
s
x
d
x
+
0
=
2
∫
π
/
2
0
(
x
2
c
o
s
x
)
d
x
=
2
{
(
x
2
s
i
n
x
)
π
/
2
0
−
∫
π
/
2
0
(
x
3
c
o
s
x
)
}
2
[
π
2
4
−
2
{
(
−
x
c
o
s
x
)
π
/
2
0
−
∫
π
/
2
0
1.
(
−
c
o
s
x
)
d
x
}
]
=
2
[
π
2
4
−
2
(
s
i
n
x
)
π
/
2
0
]
=
2
[
π
2
4
−
2
]
=
(
π
2
2
−
4
)
Let
θ
be the angle between the lines
L
1
:
[
x
=
2
t
+
1
y
=
t
+
1
z
=
3
t
+
1
and
L
2
:
[
x
=
3
s
+
2
y
=
6
s
−
1
z
=
4
where
s
,
t
∈
R
.
Then the value of
∫
θ
0
1
1
+
tan
x
d
x
=
Report Question
0%
π
/
6
0%
π
/
4
0%
π
/
2
0%
π
/
3
Let
I
1
=
∫
1
0
(
1
−
x
50
)
100
d
x
and
I
2
=
∫
1
0
(
1
−
x
50
)
101
d
x
, then
I
1
I
2
=
Report Question
0%
5051
5050
0%
5051
5049
0%
51
50
0%
101
100
∫
9
−
9
log
(
x
+
√
x
2
+
1
)
d
x
equal
Report Question
0%
2
log
(
9
2
+
1
)
0%
2
log
(
√
9
2
+
1
−
9
)
0%
0
0%
2
log
(
9
+
√
9
2
+
1
)
If
∫
π
3
0
tan
θ
√
2
k
sec
θ
d
θ
=
1
−
1
√
2
,
(
k
>
0
)
,
then the value of k is :
Report Question
0%
2
0%
1
2
0%
4
0%
1
Explanation
L
H
S
=
1
√
2
k
∫
π
3
0
tan
θ
√
sec
θ
d
θ
=
1
√
2
k
∫
π
3
0
sin
θ
√
cos
θ
d
θ
=
−
1
√
2
k
|
2
√
cos
θ
|
π
3
0
=
−
√
2
√
k
(
1
√
2
−
1
)
=
√
2
√
k
(
1
−
1
√
2
)
∵
it is given that
R
H
S
=
1
−
1
√
2
∴
by comparing we get
k
=
2
The value of
∫
2
π
0
x
sin
8
x
sin
8
x
+
cos
8
x
d
x
is equal to?
Report Question
0%
2
π
0%
π
2
0%
2
π
2
0%
4
π
Explanation
Let
I
=
∫
2
π
0
x
sin
8
x
sin
8
x
+
cos
8
x
___(1)
I
=
∫
2
π
0
(
2
π
−
x
)
sin
8
(
2
π
−
x
)
sin
8
(
2
π
−
x
)
+
cos
8
(
2
π
−
x
)
I
=
∫
2
π
0
(
2
π
−
x
)
sin
8
x
sin
8
x
+
cos
8
x
___(2)
Add (1) + (2) we get,
2
I
=
∫
2
π
0
2
π
sin
8
x
sin
8
x
+
cos
8
x
2
I
=
2
×
2
π
∫
π
0
sin
8
x
sin
8
x
+
cos
8
x
I
=
2
π
∫
π
0
sin
8
x
sin
8
x
+
cos
8
x
d
x
I
=
4
π
∫
π
/
2
0
sin
8
x
sin
8
x
+
cos
8
x
d
x
__(3)
I
=
4
π
∫
π
/
2
0
sin
8
(
π
2
−
x
)
sin
8
(
π
2
−
x
)
+
cos
8
(
π
2
−
x
)
I
=
4
π
∫
π
/
2
0
cos
8
(
x
)
sin
8
x
+
cos
8
x
__(4)
add (3) + (4) we get
I
=
2
π
∫
π
/
2
0
sin
8
x
+
cos
8
x
sin
8
x
+
cos
8
x
d
x
=
2
π
∫
π
/
2
0
1
d
x
=
2
π
×
π
2
=
π
2
If
f
(
a
−
x
)
=
−
f
(
x
)
, then
∫
a
0
f
(
x
)
d
x
=
0
.
Report Question
0%
True
0%
False
Explanation
∫
a
−
a
f
(
x
)
d
x
=
∫
0
−
a
f
(
x
)
d
x
+
∫
a
0
f
(
x
)
d
x
∫
a
a
f
(
x
)
d
x
=
I
+
∫
a
0
f
(
x
)
d
x
Now
I
=
∫
0
−
a
f
(
x
)
d
x
Put
x
=
−
t
⇒
d
x
=
−
d
t
When
x
=
−
a
,
t
=
a
and when
x
=
0
,
t
=
0
I
=
∫
0
a
f
(
−
t
)
(
−
d
t
)
=
−
∫
0
a
f
(
−
t
)
d
t
=
∫
a
0
f
(
−
t
)
d
t
.
.
.
.
.
.
.
.
.
.
.
since
∫
b
a
f
(
x
)
d
x
=
−
∫
a
b
f
(
x
)
d
x
=
∫
a
0
f
(
−
x
)
d
t
.
.
.
.
.
.
.
.
.
.
since
∫
b
a
f
(
x
)
d
x
=
−
∫
a
b
f
(
t
)
d
t
Thus,
∫
a
−
a
f
(
x
)
d
x
=
∫
a
0
f
(
−
x
)
d
x
+
∫
a
0
f
(
x
)
d
x
=
∫
a
0
[
f
(
−
x
)
+
f
(
x
)
]
d
x
=
2
∫
a
0
f
(
x
)
d
x
If
f
(
x
)
is an odd function, then
f
(
−
x
)
=
−
f
(
x
)
⇒
∫
a
−
a
f
(
x
)
d
x
=
∫
a
0
[
−
f
(
x
)
+
f
(
x
)
]
d
x
=
0
Hence proved.
Thus, the given statement is true.
∫
400
π
0
√
1
−
cos
2
x
Report Question
0%
200
√
2
0%
400
√
2
0%
800
√
2
0%
n
o
n
e
Explanation
I
=
∫
400
π
0
√
1
−
cos
x
d
x
=
400
∫
π
0
√
1
−
cos
2
x
d
x
=
800
∫
π
0
√
2
√
sin
2
x
d
x
=
800
√
2
∫
π
2
0
sin
x
d
x
=
800
√
2
[
cos
x
]
0
π
2
=
800
√
2
H
e
n
c
e
,
t
h
e
o
p
t
i
o
n
C
i
s
t
h
e
c
o
r
r
e
c
t
a
n
s
w
e
r
.
∫
1
0
sin
−
1
x
d
x
=
π
2
−
1
Report Question
0%
True
0%
False
Let a function
f
:
R
→
R
be defined as
f
(
x
)
=
x
+
sin
x
. The value of
∫
2
π
0
f
−
1
(
x
)
d
x
will
Report Question
0%
2
π
2
0%
2
π
2
−
2
0%
2
π
2
+
2
0%
π
2
Explanation
We have
f
(
x
)
=
x
+
sin
x
Let
I
=
∫
2
π
0
f
−
1
(
x
)
d
x
let
x
=
f
(
t
)
⇒
d
x
=
f
′
(
t
)
d
t
I
=
∫
f
−
1
(
2
π
)
f
−
1
(
0
)
t
.
f
′
(
t
)
d
t
=
∫
2
π
0
t
.
f
′
(
t
)
d
t
=
t
f
(
t
)
|
2
π
0
−
∫
2
π
0
f
(
t
)
d
t
=
2
π
.
f
(
2
π
)
−
∫
2
π
0
t
+
sin
t
4
π
2
−
t
2
2
|
2
π
0
+
cos
t
|
2
π
0
=
4
π
2
−
2
π
2
+
0
=
2
π
2
The value of
π
4
∫
0
tan
2
θ
d
θ
=
Report Question
0%
π
4
−
1
0%
π
4
0%
1
−
π
4
0%
none of these
Explanation
Now,
π
4
∫
0
tan
2
θ
d
θ
=
π
4
∫
0
(
sec
2
θ
−
1
)
d
θ
[ Since
sec
2
θ
−
tan
2
θ
=
1
]
=
[
tan
θ
−
θ
]
π
4
0
=
1
−
π
4
.
∫
1
0
x
(
x
2
+
1
)
3
2
d
x
=
.
.
.
.
.
.
.
.
Report Question
0%
1
3
0%
2
3
0%
3
2
0%
1
−
1
√
2
Explanation
∫
1
0
x
(
x
2
+
1
)
3
2
d
x
=
1
2
∫
1
0
d
(
x
2
+
1
)
(
x
2
+
1
)
3
/
2
=
1
2
[
(
x
2
+
1
)
−
3
/
2
+
1
−
3
/
2
+
1
]
1
0
=
[
1
2
×
(
x
2
+
1
)
−
1
/
2
−
1
/
2
]
1
0
=
−
[
1
√
2
−
1
]
=
1
−
1
√
2
∫
1
/
2
0
d
x
(
1
+
x
2
)
√
1
−
x
2
is equal to?
Report Question
0%
1
√
2
tan
−
1
√
2
3
0%
2
√
2
tan
−
1
(
3
√
2
)
0%
√
2
2
tan
−
1
(
3
2
)
0%
√
2
2
tan
−
1
(
√
3
2
)
Explanation
Let
x
=
sin
θ
d
x
=
cos
θ
d
θ
∫
π
/
6
0
cos
θ
d
θ
(
1
+
sin
2
θ
)
cos
θ
=
∫
π
/
6
0
sec
2
θ
1
+
2
tan
2
θ
=
d
θ
.
Multiply numeration and denomination by
sec
2
θ
=
1
√
2
tan
−
1
(
√
2
tan
θ
)
π
6
0
=
1
√
2
tan
−
1
(
√
2
3
)
.
∫
1
/
2
−
1
e
x
(
2
−
x
2
)
d
x
(
1
−
x
)
√
1
−
x
2
is equal to
Report Question
0%
√
e
2
(
√
3
+
1
)
0%
√
3
e
2
0%
√
3
e
0%
√
e
3
∫
3
1
4
x
x
2
+
1
d
x
Report Question
0%
log
5
0%
1
2
log
5
0%
log
25
0%
log
100
Explanation
∫
3
1
(
4
x
1
+
x
2
)
d
x
substitute
x
2
+
1
=
t
x
=
1
then
t
=
2
x
=
3
then
t
=
10
∫
10
2
2
t
d
t
=
2
[
ln
(
1
+
x
2
)
]
3
1
=
2
[
ln
10
−
ln
2
]
=
ln
25
If
∫
x
l
o
g
2
d
x
√
e
x
−
1
=
π
6
,
then x is equal to _________.
Report Question
0%
4
0%
in 8
0%
in 4
0%
None of these
∫
1
0
d
x
e
x
+
e
−
x
is equal to?
Report Question
0%
π
4
−
tan
−
1
(
e
)
0%
tan
−
1
(
e
)
−
π
4
0%
tan
−
1
(
e
)
+
π
4
0%
tan
−
1
(
e
)
Explanation
∫
1
0
d
x
e
x
+
e
−
1
=
∫
1
0
e
x
(
e
x
)
2
+
1
d
x
Put
e
x
=
t
⇒
e
x
d
x
=
d
t
x
→
0
⇒
t
→
e
0
=
1
x
→
1
⇒
t
→
e
1
=
e
∫
e
1
1
1
+
t
2
d
t
=
tan
−
1
t
+
c
=
(
tan
−
1
(
e
x
)
)
1
0
+
c
=
tan
−
1
(
e
)
−
π
4
.
If
∫
π
2
0
cot
x
cot
x
+
c
o
s
e
c
x
d
x
=
m
(
π
+
n
)
, then
m
n
is equal to?
Report Question
0%
−
1
0%
1
0%
1
2
0%
−
1
2
Explanation
∫
π
/
2
0
cot
x
d
x
cot
x
+
c
o
s
e
c
x
∫
π
/
2
0
cos
x
cos
x
+
1
=
∫
2
cos
2
x
2
−
1
2
cos
2
x
2
∫
π
/
2
0
(
1
−
1
2
sec
2
x
2
)
d
x
[
x
−
tan
x
2
]
π
2
0
⇒
1
2
[
π
−
2
]
⇒
m
=
1
2
,
n
=
−
2
m
n
=
−
1
.
Evaluate :
∫
1
0
d
x
(
1
+
x
+
x
2
)
Report Question
0%
π
√
3
0%
π
3
0%
π
3
√
3
0%
None of these
Evaluate
∫
π
/
2
0
e
x
(
1
+
sin
x
1
+
cos
x
)
d
x
Report Question
0%
0
0%
π
4
0%
e
π
/
2
0%
(
e
π
/
2
−
1
)
Explanation
Let
I
=
∫
π
/
2
0
e
x
(
1
+
sin
x
1
+
cos
x
)
d
x
=
∫
π
/
2
0
e
x
{
1
(
1
+
cos
x
)
+
sin
x
(
1
+
cos
x
)
}
d
x
=
∫
π
/
2
0
e
x
{
1
2
cos
2
(
x
/
2
)
+
2
sin
(
x
/
2
)
cos
(
x
/
2
)
2
cos
2
(
x
/
2
)
}
d
x
=
∫
π
/
2
0
e
x
{
tan
x
2
+
1
2
sec
2
x
2
}
d
x
=
[
e
x
tan
x
2
]
π
/
2
0
=
e
π
/
2
.
Evaluate :
∫
√
2
0
√
2
−
x
2
d
x
Report Question
0%
π
0%
2
π
0%
π
2
0%
None of these
Evaluate :
∫
1
0
(
1
−
x
)
(
1
+
x
)
d
x
Report Question
0%
(
log
2
+
1
)
0%
(
log
2
−
1
)
0%
(
2
log
2
−
1
)
0%
(
2
log
2
+
1
)
Evaluate :
∫
9
0
d
x
(
1
+
√
x
)
Report Question
0%
(
3
−
2
l
o
g
2
)
0%
(
3
+
2
l
o
g
2
)
0%
(
6
−
2
l
o
g
4
)
0%
(
6
+
2
l
o
g
4
)
Evaluate :
∫
2
−
2
|
x
|
d
x
Report Question
0%
4
0%
3.5
0%
2
0%
0
Evaluate
∫
1
0
(
1
−
x
)
(
1
+
x
)
d
x
Report Question
0%
1
2
l
o
g
2
0%
(
2
l
o
g
2
+
1
)
0%
(
2
l
o
g
2
−
1
)
0%
(
1
2
l
o
g
2
−
1
)
Evaluate
∫
π
/
6
0
cos
x
cos
2
x
d
x
Report Question
0%
1
4
0%
5
12
0%
1
3
0%
7
12
The value of
∫
199
π
/
2
−
π
/
2
√
(
1
+
cos
2
x
)
d
x
is?
Report Question
0%
50
√
2
0%
100
√
2
0%
150
√
2
0%
200
√
2
Evaluate :
∫
2
1
|
x
2
−
3
x
+
2
|
d
x
Report Question
0%
−
1
6
0%
1
6
0%
1
3
0%
2
3
Given
I
m
=
∫
e
1
(
log
x
)
m
d
x
.
If
I
m
K
+
I
m
−
2
L
=
e
,
then the values of
K
and
L
are
Report Question
0%
1
1
−
m
,
1
m
0%
(
1
−
m
)
,
1
m
0%
1
1
−
m
,
m
(
m
−
2
)
m
−
1
0%
m
m
−
1
,
m
−
2
Explanation
I
m
=
∫
e
1
(
log
x
)
m
d
x
∴
I
m
=
[
x
(
log
x
)
m
]
e
1
−
∫
e
1
x
m
(
log
x
)
m
−
1
x
d
x
(integrating by parts)
⇒
˙
I
m
=
e
−
m
∫
e
1
(
log
x
)
m
−
1
d
x
=
e
−
m
I
m
−
1
Replacing
m
by
m
−
1
I
m
−
1
=
e
−
(
m
−
1
)
I
m
−
2
From equations
(
1
)
and
(
2
)
,
we have
I
m
=
e
−
m
[
e
−
(
m
−
1
)
I
m
−
2
]
⇒
I
m
−
m
(
m
−
1
)
I
m
−
1
=
e
(
1
−
m
)
⇒
I
m
1
−
m
+
m
I
m
−
2
=
e
⇒
K
=
1
−
m
and
L
=
1
m
∫
a
−
a
x
|
x
|
d
x
=
?
Report Question
0%
0
0%
2
a
0%
2
a
3
3
0%
None of these
Explanation
|
x
|
=
−
x
,
x
<
0
x
,
x
>
0
I
=
∫
0
−
a
x
(
−
x
)
d
x
+
∫
a
0
x
⋅
x
d
x
=
∫
0
−
a
−
x
2
d
x
+
∫
a
0
x
2
d
x
=
[
−
x
3
3
]
0
−
a
+
[
x
3
3
]
a
0
=
+
1
3
(
−
a
)
3
+
a
3
3
=
0
.
∫
1
−
2
|
x
|
2
d
x
=
?
Report Question
0%
3
0%
2.5
0%
1.5
0%
None of these
Explanation
|
x
|
=
−
x
,
x
<
0
x
,
x
>
0
I
=
∫
0
−
2
−
x
x
d
x
+
∫
1
0
x
x
d
x
=
∫
0
−
2
−
d
x
+
∫
1
0
d
x
=
[
−
x
]
0
−
2
+
[
x
]
1
0
=
(
−
2
+
1
)
=
−
1
.
Evaluate :
∫
1
0
|
2
x
−
1
|
d
x
Report Question
0%
2
0%
1
2
0%
1
0%
0
Value of
∫
3
2
d
x
√
(
1
+
x
3
)
is?
Report Question
0%
Less than
1
0%
Greater than
2
0%
Lies between
3
and
4
0%
None of these
Explanation
Given,
∫
3
2
d
x
√
1
+
x
3
Let us consider the function
f
(
x
)
=
∫
3
2
d
x
√
1
+
x
3
according to the question putting the lower and upper limit we get,
f
(
2
)
=
1
√
1
+
2
3
=
1
3
f
(
3
)
=
1
√
1
+
3
3
=
1
√
28
∴
1
3
>
1
√
28
we conclude that the Larger integrating value lies in the point 2
∴
1
√
28
<
f
(
x
)
<
1
3
Since, m < f(x) < M
⇒
∫
m
d
x
<
∫
f
(
x
)
d
x
<
∫
M
d
x
[
∀
m
,
M
∈
R
and be the lower and upper limit of the f(n) ]
we can conclude that,
∫
3
2
f
(
x
)
d
x
<
1
3
Therefore the value is less than 1. (Ans)
The value of the definite integral
∫
π
/
2
0
sin
5
x
sin
x
d
x
is
Report Question
0%
0
0%
π
2
0%
π
0%
2
π
Explanation
sin
n
x
−
sin
(
n
−
2
)
x
=
2
cos
(
n
−
1
)
x
sin
x
⇒
∫
sin
n
x
sin
x
d
x
=
∫
2
cos
(
n
−
1
)
d
x
+
∫
sin
(
n
−
2
)
x
sin
x
d
x
∴
∫
π
/
2
0
sin
5
x
sin
x
d
x
=
∫
π
/
2
0
2
cos
4
x
d
x
+
∫
π
/
2
0
sin
3
x
sin
x
d
x
=
0
+
∫
π
/
2
0
sin
3
x
sin
x
d
x
=
∫
π
/
2
0
3
−
4
s
i
n
2
x
d
x
=
∫
π
/
2
0
1
+
2
c
o
s
2
x
d
x
=
π
2
Evaluate :
∫
1
−
2
|
2
x
+
1
|
d
x
Report Question
0%
5
2
0%
7
2
0%
9
2
0%
4
The value of the definite integral
∫
4
2
(
x
(
3
−
x
)
(
4
+
x
)
(
6
−
x
)
(
10
−
x
)
+
sin
x
)
d
x
equals
Report Question
0%
cos
2
+
cos
4
0%
cos
2
−
cos
4
0%
sin
2
+
sin
4
0%
sin
2
−
sin
4
Explanation
Let
I
=
∫
4
2
(
x
(
3
−
x
)
(
4
+
x
)
(
6
−
x
)
(
10
−
x
)
+
sin
x
)
d
x
)
.........(i)
=
∫
4
2
(
(
6
−
x
)
(
3
−
(
6
−
x
)
)
(
4
+
(
6
−
x
)
)
(
6
−
(
6
−
x
)
)
..........
∵
∫
b
a
f
(
x
)
d
x
=
∫
b
a
f
(
a
+
b
−
x
)
d
x
=
∫
4
2
(
(
6
−
x
)
(
x
−
3
)
(
10
−
x
)
x
(
4
+
x
)
+
sin
(
6
−
x
)
)
d
x
.......... (ii)
Adding equations (1) and
(
2
)
,
we get
2
I
=
∫
4
2
(
sin
x
+
sin
(
6
−
x
)
)
d
x
=
(
−
cos
x
+
cos
(
6
−
x
)
)
4
2
=
−
cos
4
+
cos
2
+
cos
2
−
cos
4
=
2
(
cos
2
−
cos
4
)
⇒
I
=
cos
2
−
cos
4
Let
f
:
R
→
R
be a function as
f
(
x
)
=
(
x
−
1
)
(
x
+
2
)
(
x
−
3
)
(
x
−
6
)
−
100
. If
g
(
x
)
is a polynomial of degree
≤
3
such that
∫
g
(
x
)
f
(
x
)
d
x
does not contain any logarithm function and
g
(
−
2
)
=
10
. Then
∫
g
(
x
)
f
(
x
)
d
x
, equals
Report Question
0%
tan
−
1
(
x
−
2
2
)
+
c
0%
tan
−
1
(
x
−
1
1
)
+
c
0%
tan
−
1
(
x
)
+
c
0%
None of these
Explanation
Here,
f
(
x
)
=
(
x
−
1
)
(
x
+
2
)
(
x
−
3
)
(
x
−
6
)
−
100
=
(
x
2
+
4
x
+
3
)
(
x
2
−
4
x
−
12
)
−
100
=
(
x
2
−
4
x
)
2
−
9
(
x
2
−
4
x
)
−
136
=
(
x
2
−
4
x
+
8
)
(
x
2
−
4
x
+
17
)
∫
g
(
x
)
f
(
x
)
=
g
(
x
)
(
x
2
−
4
x
−
17
)
(
x
2
−
4
x
+
8
)
=
A
x
+
B
x
2
−
4
x
−
17
+
C
x
+
D
x
2
−
4
x
+
8
Clearly,
A
,
B
and
C
must be zero.
∴
g
(
x
)
(
x
2
−
4
x
−
17
)
(
x
2
−
4
x
+
8
)
=
D
x
2
−
4
x
+
8
∴
g
(
x
)
=
D
(
x
2
−
4
x
−
17
)
g
(
−
2
)
=
D
(
4
+
8
−
17
)
=
−
10
[given]
⇒
g
(
x
)
f
(
x
)
=
2
(
x
2
−
4
x
−
17
)
(
x
2
−
4
x
−
17
)
(
x
2
−
4
x
+
8
)
=
2
x
2
−
4
x
+
8
∴
∫
g
(
x
)
f
(
x
)
d
x
=
∫
2
x
2
−
4
x
+
8
d
x
=
2
∫
d
x
(
x
−
2
)
2
+
(
2
)
2
=
2.
1
2
tan
−
1
(
x
−
2
2
)
+
C
=
tan
−
1
(
x
−
2
2
)
+
C
If
∫
−
1
−
2
(
a
x
2
−
5
)
d
x
and
5
+
∫
2
1
(
b
x
+
c
)
d
x
=
0
,
then
Report Question
0%
a
x
2
−
b
x
+
c
=
0
has atleast one root in (1,2)
0%
a
x
2
−
b
x
+
c
=
0
has atleast one root in (-2,-1)
0%
a
x
2
+
b
x
+
c
=
0
has atleast one root in (-2,-1)
0%
None of the above
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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