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CBSE Questions for Class 12 Commerce Maths Integrals Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Integrals
Quiz 11
What is
∫
3
1
|
1
−
x
4
|
d
x
equal to
?
Report Question
0%
−
232
/
5
0%
−
116
/
5
0%
116
/
5
0%
232
/
5
The value of
∫
2
3
/
2
(
x
−
1
3
−
x
)
1
/
2
d
x
is
Report Question
0%
√
3
2
−
1
+
π
6
0%
√
3
2
−
1
+
π
8
0%
√
3
2
+
1
−
π
6
0%
None of these
The value of integral
∫
b
a
|
x
|
x
d
x
,
a
<
b
is
Report Question
0%
|
a
|
−
|
b
|
0%
|
b
|
−
|
a
|
0%
|
a
|
−
b
0%
|
b
|
−
a
The value of the definite integral
∫
1
0
(
1
+
e
−
x
2
)
d
x
is
Report Question
0%
−
1
0%
2
0%
1
+
e
−
1
0%
None of the above
Let
J
=
∫
∞
0
I
n
x
1
+
x
3
d
x
and
K
=
∫
∞
0
x
I
n
x
1
+
x
3
d
x
then
Report Question
0%
J
+
K
=
0
0%
J
−
K
=
0
0%
J
+
K
<
0
0%
J
+
K
>
0
The integral
∫
π
/
3
π
/
6
f
(
x
)
f
(
x
)
+
f
(
π
2
−
x
)
d
x
is equal to
Report Question
0%
π
12
0%
12
π
0%
π
13
0%
13
π
If
a
>
0
and
A
=
∫
a
0
cos
−
1
x
d
x
, then
∫
a
−
a
(
cos
−
1
x
−
sin
1
√
1
−
x
2
)
d
x
=
π
a
−
λ
A
. Then
λ
Report Question
0%
0
0%
3
0%
2
0%
n
o
n
e
o
f
t
h
e
s
e
∫
π
/
2
π
/
4
√
2
+
√
2
+
2
cos
4
x
d
x
is equal to
Report Question
0%
√
2
0%
√
2
(
√
2
−
1
)
0%
2
0%
n
o
n
e
o
f
t
h
e
s
e
∫
(
1
+
x
4
)
1
4
d
x
=
A
I
n
(
(
1
+
x
4
)
1
4
+
x
(
1
+
x
4
)
1
4
−
x
)
+
B
tan
−
1
(
(
1
+
x
4
)
1
4
x
)
+
C
,
t
h
e
n
A
+
B
=
Report Question
0%
1/4
0%
1/2
0%
−
1
4
0%
−
1
2
∫
5
4
e
(
x
+
5
)
2
d
x
+
3
∫
2
/
3
1
/
3
e
9
(
x
2
3
)
2
d
x
is equal to
Report Question
0%
1
0%
-1
0%
0
0%
2
Let
1
n
=
∫
1
2
0
1
√
1
−
x
n
d
x
where
n
>
2
,
then
Report Question
0%
I
n
<
π
6
0%
I
n
>
π
6
0%
I
n
<
1
2
0%
I
n
>
1
2
If
∫
b
−
c
0
f
(
x
+
c
)
d
x
=
k
∫
c
b
f
(
x
)
d
x
then k=
Report Question
0%
0
0%
1
0%
2
0%
-1
If
l
1
−
∫
1
0
2
x
2
,
l
2
−
∫
1
0
2
x
3
d
x
,
l
3
−
∫
2
1
2
x
2
d
x
and
l
4
−
∫
2
1
2
x
3
d
x
then
Report Question
0%
l
2
>
l
1
0%
l
1
>
l
2
0%
l
3
−
l
2
0%
None of these
Find the value of the equation :
∫
ln
(
1
λ
)
ln
λ
f
(
x
2
3
)
(
f
(
x
)
+
f
(
−
x
)
)
g
(
3
x
2
)
(
g
(
x
)
−
g
(
−
x
)
)
d
x
=
?
where
λ
>
1
Report Question
0%
0
0%
1
0%
λ
0%
1
/
λ
Find the value of the equation
∫
∞
0
d
x
(
x
+
√
x
2
+
1
)
3
=
?
Report Question
0%
3
/
8
0%
1
/
8
0%
−
3
/
8
0%
−
1
/
8
7
(
∫
0
π
x
4
(
1
−
x
)
4
d
x
1
+
x
2
+
π
)
is equal to
Report Question
0%
21
0%
22
0%
23
0%
n
o
n
e
o
f
t
h
e
s
e
∫
1
0
√
1
−
x
1
+
x
d
x
, is equal to:
Report Question
0%
π
4
−
1
0%
π
2
−
1
0%
π
4
+
1
0%
π
2
+
1
∫
1
0
√
1
−
x
1
+
x
d
x
,
is equal to
Report Question
0%
π
4
−
1
0%
π
2
−
1
0%
π
4
+
1
0%
π
2
+
1
1
π
∫
2
−
2
1
4
+
x
2
d
x
=
Report Question
0%
0
0%
1
4
0%
π
4
0%
1
2
∫
8
0
[
√
t
]
d
t
at equals to (where [.] greatest integer function.)
Report Question
0%
28
0%
11
0%
2
0%
8
If
f
(
x
)
=
∫
1
0
(
x
f
(
t
)
+
1
)
d
t
,
t
h
e
n
∫
3
0
f
(
x
)
d
x
=
12
because
Statement-2: f(x) = 3x + 1
Report Question
0%
Statements-1 is true, statements-2 is true and statements-2 is correct explanation for statement-1.
0%
Statements-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statements-1.
0%
Statements-1 is true, statements-2 is false.
0%
Statements-1 is false, statements-2 is true.
If
I
m
=
e
∫
1
(
l
n
x
)
m
d
x
,
where
m
ϵ
N
,
then
I
10
+
10
I
9
is equal to-
Report Question
0%
e
10
0%
e
10
10
0%
e
0%
e-1
If for every integer n,
∫
n
+
1
n
f
(
x
)
d
x
=
n
2
, then the value of
∫
4
−
2
f
(
x
)
d
x
is -
Report Question
0%
16
0%
14
0%
19
0%
None of these.
E
=
∫
∞
R
G
M
m
x
2
dx , (where
G
,
M
,
m
are constants ) equal to
Report Question
0%
−
G
M
m
R
2
0%
+
G
M
m
R
2
0%
−
G
M
m
R
0%
+
G
M
m
R
∫
4036
0
2
x
2
x
+
1
4036
−
x
d
x
=
.
.
.
.
.
.
.
.
.
.
.
.
Report Question
0%
2018
0%
4035
0%
2017
0%
-2015
If
I
1
=
∫
1
x
1
1
+
t
2
d
t
and
I
2
=
∫
1
/
x
1
1
1
+
t
2
d
t
for x > 0, then
Report Question
0%
I
1
=
I
2
0%
I
1
>
I
2
0%
I
2
>
I
1
0%
None of these.
The value of the integral
∫
1
1
3
(
x
−
x
3
)
1
3
x
4
d
x
is
Report Question
0%
6
0%
0
0%
3
0%
4
−
5
∫
−
4
e
(
x
+
5
)
2
d
x
+
3
2
/
3
∫
1
/
3
e
9
(
9
(
x
−
2
/
3
)
2
dx is equal toi
Report Question
0%
e
5
0%
e
4
0%
3
e
2
0%
0
If
z
=
x
+
3
i
then value of
∫
4
2
[
a
r
g
|
z
−
i
z
+
i
|
]
d
x
, where
[
.
]
denotes the greatest integer function, is?
Report Question
0%
3
√
2
0%
6
√
3
0%
√
6
0%
None
The area of the region bounded by the lines
x
=
1
,
x
=
2
, and the curves
x
(
y
−
e
x
)
=
sin
x
and
2
x
y
=
2
sin
x
+
x
3
is
Report Question
0%
e
2
−
e
−
1
6
0%
e
2
−
e
−
7
6
0%
e
2
−
e
+
1
6
0%
e
2
−
e
+
7
6
Explanation
Given:
x
(
y
−
e
x
)
=
sin
x
⇒
x
y
−
x
e
x
=
sin
x
⇒
x
y
=
sin
x
+
x
e
x
⇒
y
=
sin
x
x
+
e
x
Given:
2
x
y
=
2
sin
x
+
x
3
⇒
y
=
sin
x
x
+
x
2
2
A
=
∫
2
1
{
sin
x
x
+
e
x
−
(
sin
x
x
+
x
2
2
)
}
d
x
=
∫
2
1
(
e
x
−
x
2
2
)
d
x
=
[
e
x
−
x
3
6
]
2
1
=
e
2
−
8
6
−
(
e
−
1
6
)
=
e
2
−
e
−
7
6
Suppose
I
1
=
∫
π
/
2
0
cos
(
π
sin
2
x
)
d
x
;
I
2
=
∫
π
/
2
0
cos
(
2
π
sin
2
x
)
d
x
and
I
3
=
∫
π
/
2
0
cos
(
π
sin
x
)
d
x
then
Report Question
0%
I
1
=
0
0%
I
2
+
I
3
=
0
0%
I
1
+
I
2
+
I
3
=
0
0%
I
2
=
I
3
Explanation
We have
I
1
=
∫
π
/
2
0
c
o
s
(
π
s
i
n
2
x
)
d
x
using
∫
b
a
f
(
x
)
d
x
=
∫
b
a
f
(
a
+
b
−
x
)
d
x
I
1
=
∫
π
/
2
0
c
o
s
(
π
sin
2
(
π
2
−
x
)
)
d
x
I
1
=
∫
π
/
2
0
c
o
s
(
π
c
o
s
2
x
)
d
x
On adding,
2
I
1
=
∫
π
/
2
0
c
o
s
(
π
s
i
n
2
x
)
+
c
o
s
(
π
c
o
s
2
x
)
d
x
=
∫
π
/
2
0
2
c
o
s
(
π
2
)
.
c
o
s
(
π
2
c
o
s
2
x
)
d
x
⇒
I
1
=
0
Now,
I
2
=
∫
π
/
2
0
cos
(
2
π
sin
2
x
)
d
x
I
2
=
∫
π
/
2
0
c
o
s
{
π
(
1
−
cos
2
x
)
}
d
x
=
−
∫
π
/
2
0
cos
(
π
cos
2
x
)
d
x
=
−
1
2
∫
π
/
4
0
c
o
s
(
π
c
o
s
t
)
d
t
(
p
u
t
2
x
=
t
)
=
−
2
2
∫
π
/
2
0
c
o
s
(
π
c
o
s
t
)
d
t
=
−
I
3
⇒
I
2
+
I
3
=
0
I
3
=
−
∫
π
/
2
0
c
o
s
(
π
s
i
n
t
)
d
t
∴
I
2
+
I
3
=
0
Hence,
I
1
+
I
2
+
I
3
=
0
If
∫
cos
x
d
x
sin
3
x
(
1
+
sin
6
x
)
2
/
3
=
f
(
x
)
(
1
+
sin
6
x
)
1
/
α
+
c
Where
c
is a constant of integration, then
λ
f
(
π
3
)
is equal to:
Report Question
0%
9
8
0%
−
2
0%
2
0%
−
9
8
Explanation
∫
cos
x
d
x
sin
3
x
(
1
+
sin
6
x
)
2
/
3
=
f
(
x
)
(
1
+
sin
6
x
)
1
/
α
+
c
.
.
.
.
.
g
i
v
e
n
Let
sin
x
=
t
cot
x
d
x
=
d
t
I
=
∫
d
t
t
3
(
1
+
t
6
)
2
/
3
I
=
∫
d
t
t
3
.
t
4
(
1
+
1
t
6
)
2
/
3
I
=
∫
d
t
t
7
(
1
+
1
t
6
)
2
/
3
Put
1
+
1
t
6
=
r
3
⇒
d
t
t
7
=
−
1
2
r
2
d
r
=
−
1
2
∫
r
2
d
r
r
2
=
−
1
2
r
+
c
=
−
1
2
(
sin
6
x
+
1
sin
6
x
)
1
/
3
+
c
.
.
.
.
.
.
.
As
r
=
(
1
+
1
t
6
)
1
/
3
and
t
=
sin
x
=
−
1
2
sin
2
x
(
1
+
sin
6
x
)
1
/
3
+
c
f
(
x
)
=
−
1
2
c
o
s
e
c
2
x
and
λ
=
3
Now for
x
=
π
3
c
o
s
e
c
π
3
=
2
√
3
λ
f
(
π
3
)
=
−
1
2
×
(
2
√
3
)
2
×
3
λ
f
(
π
3
)
=
−
2
.
.
.
.
.
A
n
s
w
e
r
Hence option
′
B
′
is the answer.
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