Processing math: 100%
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 4 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 4
If
y
=
l
o
g
x
3
+
3
s
i
n
−
1
x
+
k
x
2
, then find
d
y
d
x
Report Question
0%
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
3
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
−
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
2
x
Explanation
Here,
y
=
log
x
3
+
3
sin
−
1
x
+
k
x
2
On differentiating we get
d
y
d
x
=
d
d
x
[
log
x
3
]
+
d
d
x
[
3
sin
−
1
x
]
+
d
d
x
[
k
x
2
]
=
3
d
d
x
[
log
x
]
+
3
d
d
x
(
sin
−
1
x
)
+
k
d
d
x
(
x
2
)
=
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
If
y
=
5
3
−
x
2
+
(
3
−
x
2
)
5
, then
d
y
d
x
=
Report Question
0%
−
2
x
{
5
3
−
x
2
⋅
log
e
5
+
5
(
3
−
x
2
)
4
}
0%
−
x
{
5
3
−
x
2
⋅
log
e
5
+
5
(
3
−
x
2
)
4
}
0%
−
2
x
{
5
3
−
x
2
⋅
log
e
5
+
(
3
−
x
2
)
4
}
0%
−
2
x
{
5
3
−
x
2
+
5
(
3
−
x
2
)
4
}
Explanation
d
d
x
(
5
3
−
x
2
+
(
3
−
x
2
)
5
)
=
5
3
−
x
2
log
e
5
d
d
x
(
3
−
x
2
)
+
5
(
3
−
x
2
)
4
d
d
x
(
3
−
x
2
)
=
5
3
−
x
2
log
e
5
(
−
2
x
)
+
5
(
3
−
x
2
)
4
(
−
2
x
)
=
−
2
x
(
5
3
−
x
2
log
e
5
+
5
(
3
−
x
2
)
)
If
y
=
e
a
x
⋅
cos
(
b
x
+
c
)
, then find
d
y
d
x
Report Question
0%
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
0%
a
e
a
x
cos
(
b
x
+
c
)
+
b
e
a
x
sin
(
b
x
+
c
)
0%
e
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
0%
a
e
a
x
cos
(
b
x
+
c
)
Explanation
y
=
e
a
x
⋅
cos
(
b
x
+
c
)
On differentiating, we get
d
y
d
x
=
e
a
x
d
d
x
(
cos
(
b
x
+
c
)
)
+
cos
(
b
x
+
c
)
d
d
x
e
a
x
=
−
e
a
x
sin
(
b
x
+
c
)
d
d
x
(
b
x
+
c
)
+
cos
(
b
x
+
c
)
e
a
x
d
d
x
(
a
x
)
=
−
b
e
a
x
sin
(
b
x
+
c
)
+
a
cos
(
b
x
+
c
)
e
a
x
=
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
If
y
=
log
3
x
+
3
log
e
x
+
2
tan
x
, then
d
y
d
x
=
Report Question
0%
1
x
log
e
3
+
3
x
+
2
sec
2
x
0%
1
x
log
e
3
+
3
x
+
sec
2
x
0%
1
log
e
3
+
3
x
+
2
sec
2
x
0%
1
x
log
e
3
−
3
x
+
2
sec
2
x
Explanation
d
d
x
(
log
3
x
+
3
log
e
x
+
2
tan
x
)
=
d
d
x
(
log
e
x
log
3
+
3
log
e
x
+
2
tan
x
)
=
1
x
log
e
3
+
3
x
+
2
sec
2
x
If
y
=
e
x
log
a
+
e
a
log
x
+
e
a
log
a
, then
d
y
d
x
=
Report Question
0%
a
x
log
a
+
x
a
−
1
0%
a
x
log
a
+
a
x
0%
a
x
log
a
+
a
x
a
−
1
0%
a
x
log
a
+
a
x
a
Explanation
Let
y
=
e
x
l
o
g
a
+
e
a
l
o
g
x
+
e
a
l
o
g
a
=
e
log
e
a
x
+
e
log
e
x
a
+
e
log
e
a
a
=
a
x
+
x
a
+
a
a
......... Since
e
log
e
x
=
x
d
y
d
x
=
d
d
x
(
a
x
+
x
a
+
a
a
)
=
a
x
log
a
+
a
.
x
a
−
1
If
y
=
1
1
+
x
β
−
α
+
x
γ
−
α
+
1
1
+
x
α
−
β
+
x
γ
−
β
+
1
1
+
x
α
−
γ
+
x
β
−
γ
then
d
y
d
x
is equal to-
Report Question
0%
0
0%
1
0%
(
a
+
β
+
γ
)
X
α
+
β
+
γ
−
1
0%
None of these
If
2
x
+
2
y
=
2
x
+
y
, then
d
y
d
x
has the value equal to
Report Question
0%
−
2
y
2
x
0%
1
1
−
2
x
0%
1
−
2
y
0%
2
x
(
1
−
2
y
)
2
y
(
2
x
−
1
)
Explanation
2
x
+
2
y
=
2
x
+
y
Differentiating both the sides
⇒
d
d
x
(
2
x
+
2
y
)
=
d
d
x
(
2
x
+
y
)
⇒
2
x
log
2
+
2
y
log
2
d
y
d
x
=
2
x
+
y
log
2
(
1
+
d
y
d
x
)
⇒
log
2
(
2
x
+
2
y
d
y
d
x
)
=
log
2
(
2
x
+
y
+
2
x
+
y
d
y
d
x
)
⇒
(
2
y
−
2
x
+
y
)
d
y
d
x
=
2
x
+
y
−
2
x
⇒
d
y
d
x
=
2
x
2
y
−
2
x
2
y
−
2
x
2
y
=
2
x
(
2
y
−
1
)
2
y
(
1
−
2
x
)
=
2
x
(
1
−
2
y
)
2
y
(
2
x
−
1
)
If
f
′
(
x
)
=
sin
x
+
sin
4
x
⋅
cos
x
, then
f
′
(
2
x
2
+
π
2
)
is
Report Question
0%
4
x
{
cos
(
2
x
2
)
−
s
i
n
8
x
2
⋅
sin
2
x
2
}
0%
4
x
{
cos
(
2
x
2
)
+
sin
8
x
2
⋅
sin
2
x
2
}
0%
{
cos
(
2
x
2
)
−
sin
8
x
⋅
sin
2
x
2
}
0%
none of the above
Explanation
f
′
(
x
)
=
sin
x
+
sin
4
x
⋅
cos
x
f
′
(
2
x
2
+
π
2
)
=
d
d
x
(
2
x
2
+
π
2
)
[
sin
(
2
x
2
+
π
2
)
+
sin
4
(
2
x
2
+
π
2
)
.
cos
(
2
x
2
+
π
2
)
]
=
4
x
[
cos
2
x
2
−
sin
8
x
2
.
sin
2
x
2
]
If
y
=
|
cos
x
|
+
|
sin
x
|
, then
d
y
d
x
at
x
=
2
π
3
is
Report Question
0%
1
2
(
√
3
+
1
)
0%
2
(
√
3
−
1
)
0%
1
2
(
√
3
−
1
)
0%
none of these
Explanation
d
y
d
x
=
d
d
x
(
|
c
o
s
x
|
+
|
s
i
n
x
|
)
=
−
sin
x
+
cos
x
at
x
=
2
π
3
=
|
−
sin
2
π
3
|
+
|
cos
2
π
3
|
=
√
3
2
−
1
2
=
1
2
(
√
3
−
1
)
Find the derivative of
e
x
sin
x
Report Question
0%
e
x
sin
x
(
x
cos
x
−
sin
x
)
0%
e
x
sin
x
x
cos
x
0%
e
x
sin
x
(
−
x
cos
x
+
sin
x
)
0%
e
x
sin
x
(
x
cos
x
+
sin
x
)
Explanation
Find the derivative of
sec
−
1
(
x
+
1
x
−
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
Report Question
0%
0
0%
1
0%
−
1
0%
x
+
1
x
−
1
Explanation
Let
y
=
sec
−
1
(
x
+
1
x
−
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
y
=
cos
−
1
(
x
−
1
x
+
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
=
π
2
∴
d
y
d
x
=
d
d
x
(
π
2
)
=
0
∵
sin
−
1
θ
+
cos
−
1
θ
=
π
2
If
y
=
log
10
x
+
log
x
10
+
log
x
x
+
log
10
10
, then
d
y
d
x
=
Report Question
0%
1
x
log
e
10
−
log
e
10
x
(
log
e
x
)
2
0%
1
log
e
10
−
log
e
10
x
(
log
e
x
)
2
0%
1
x
log
e
10
−
log
e
10
x
2
(
log
e
x
)
2
0%
None of these
Explanation
d
d
x
(
log
10
x
+
log
x
10
+
log
x
x
+
log
10
10
)
=
1
x
log
e
10
+
(
−
log
e
10
(
1
x
)
)
(
log
e
x
)
2
=
1
x
log
e
10
−
log
e
10
x
(
log
e
x
)
2
If
y
=
x
a
+
x
b
+
x
a
+
x
b
+
.
.
.
.
.
∞
, then
d
y
d
x
=
Report Question
0%
a
a
b
+
2
a
y
0%
b
a
b
+
2
b
y
0%
a
a
b
+
2
b
y
0%
b
a
b
+
2
a
y
Explanation
y
=
x
a
+
x
b
+
y
⇒
y
=
x
(
b
+
y
)
a
b
+
a
y
+
x
⇒
a
b
y
+
a
y
2
+
x
y
=
b
x
+
x
y
⇒
a
b
y
+
a
y
2
=
b
x
Differentiating both sides w.r.t
x
⇒
(
a
b
+
2
a
y
)
d
y
d
x
=
b
∴
d
y
d
x
=
b
a
b
+
2
a
y
If
y
=
sin
x
1
+
cos
x
1
+
sin
x
1
+
cos
x
1
+
sin
x
1
+
.
.
.
.
.
∞
,
then
y
′
(
0
)
is
Report Question
0%
equal to
0
0%
equal to
1
2
0%
equal to
1
0%
non existent
Explanation
y
=
sin
x
1
+
cos
x
1
+
sin
x
1
+
cos
x
1
+
sin
x
1
+
.
.
.
.
.
∞
here,
y
(
0
)
=
0
⇒
y
=
sin
x
1
+
cos
x
1
+
y
⇒
y
(
1
+
y
+
cos
x
)
=
(
1
+
y
)
sin
x
differentiating wrt x
y
′
(
1
+
y
+
cos
x
)
+
y
(
y
′
−
sin
x
)
=
(
1
+
y
)
cos
x
+
y
′
sin
x
y
′
(
0
)
(
1
+
y
(
0
)
+
1
)
+
y
(
0
)
(
y
′
(
0
)
)
=
(
1
+
y
(
0
)
)
y
′
(
0
)
=
1
2
Given :
f
(
x
)
=
4
x
3
−
6
x
2
cos
2
a
+
3
x
sin
2
a
.
sin
6
a
+
√
ln
(
2
a
−
a
2
)
then
Report Question
0%
f
(
x
)
is not defined at
x
=
1
2
0%
f
′
(
1
2
)
<
0
0%
f
′
(
x
)
is not defined at
x
=
1
2
0%
f
′
(
1
2
)
>
0
Explanation
f
′
(
x
)
=
12
x
2
−
12
x
c
o
s
2
a
+
3
s
i
n
2
a
.
s
i
n
6
a
+
0
f
′
(
1
/
2
)
=
12
4
−
12
2
c
o
s
2
a
+
3
s
i
n
2
a
.
s
i
n
6
a
=
3
−
6
c
o
s
2
a
+
1.5
(
c
o
s
4
a
−
c
o
s
8
a
)
So, it will alwyas be greater than
0
If
y
=
s
e
c
−
1
(
√
x
+
1
√
x
−
1
)
+
sin
−
1
(
√
x
−
1
√
x
+
1
)
, then
d
y
d
x
equals
Report Question
0%
1
0%
0
0%
√
x
+
1
√
x
−
1
0%
√
x
−
1
√
x
+
1
Explanation
y
=
s
e
c
−
1
(
√
x
+
1
√
x
−
1
)
+
sin
−
1
(
√
x
−
1
√
x
+
1
)
y
=
cos
−
1
(
√
x
−
1
√
x
+
1
)
+
sin
−
1
(
√
x
−
1
√
x
+
1
)
=
π
2
.............
∵
sin
−
1
θ
+
cos
−
1
θ
=
π
2
∴
d
y
d
x
=
d
d
x
(
π
2
)
=
0
Which of the following could be the sketch graph of
y
=
d
d
x
(
x
ln
x
)
Report Question
0%
0%
0%
0%
Explanation
y
=
d
d
x
(
x
ln
x
)
=
ln
x
+
x
x
=
ln
x
+
1
At
x
=
0
,
y
=
1
At
y
=
0
,
x
=
e
−
1
The solution set of
f
′
(
x
)
>
g
′
(
x
)
where
f
(
x
)
=
1
2
(
5
2
x
+
1
)
&
g
(
x
)
=
5
x
+
4
x
(
ln
5
)
is
Report Question
0%
x
>
1
0%
0
<
x
<
1
0%
x
≤
0
0%
x
>
0
Explanation
Given,
f
(
x
)
=
1
2
(
5
2
x
+
1
)
f
′
(
x
)
=
5
2
x
+
1
×
log
5
g
′
(
x
)
=
5
x
l
log
5
+
4
ln
5
=
ln
5
(
5
x
+
4
)
f
′
(
x
)
>
g
′
(
x
)
=>
5
×
5
2
x
>
(
5
x
+
4
)
Let
t
=
5
x
=>
5
t
2
−
t
−
4
>
0
=>
(
t
−
1
)
(
t
+
4
5
)
>
0
=>
t
>
1
or
t
>
−
4
5
=>
5
x
>
5
0
=>
x
>
0
so,
x
>
0
The equation
y
2
e
x
y
=
9
e
−
3
.
x
2
defines
y
as a differentiable function of x. The value of
d
y
d
x
for
x
=
−
1
and
y
=
3
is
Report Question
0%
−
15
2
0%
−
9
5
0%
3
0%
15
Explanation
y
2
e
x
y
=
9
e
−
3
.
x
2
Differentiate both sides
d
d
x
(
y
2
e
x
y
)
=
d
d
x
(
9
e
−
3
x
2
)
2
y
d
y
d
x
e
x
y
+
y
2
e
x
y
(
x
d
y
d
x
+
y
)
=
9
e
−
3
(
2
x
)
On solving, we get
d
y
d
x
=
18
x
e
−
3
−
y
3
e
x
y
2
y
e
x
y
+
y
2
e
x
y
x
putting
x
=
−
1
,
y
=
3
, we get
d
y
d
x
=
−
18
e
−
3
−
27
e
−
3
6
e
−
3
−
9
e
−
3
=
−
45
e
−
3
−
3
e
−
3
=
15
f
:
R
→
R
and
f
(
x
)
=
x
(
x
4
+
1
)
(
x
+
1
)
+
x
4
+
2
x
2
+
x
+
1
, then
f
(
x
)
is
Report Question
0%
one-one ito
0%
many-one onto
0%
one-one onto
0%
many-one into
Explanation
f
(
x
)
=
x
(
x
4
+
1
)
(
x
+
1
)
+
x
4
+
2
x
2
+
x
+
1
=
x
(
x
4
+
1
)
(
x
+
1
)
+
(
x
4
+
1
)
+
1
x
2
+
x
+
1
=
(
x
4
+
1
)
(
x
2
+
x
+
1
)
+
1
x
2
+
x
+
1
=
(
x
4
+
1
)
+
1
x
2
+
x
+
1
f
′
(
x
)
=
4
x
3
−
2
x
+
1
(
x
2
+
x
+
1
)
2
=
not always positive or negative
Thus,
f
is many one.
Also range and co-domain of
f
are not same,
Hence is many-one into function
Suppose the function
f
(
x
)
−
f
(
2
x
)
has the derivative
5
at
x
=
1
and derivative
7
at
x
=
2
.The derivative of the function
f
(
x
)
−
f
(
4
x
)
at
x
=
1
, has the value equal to
Report Question
0%
19
0%
9
0%
17
0%
14
If
y
=
x
1
/
x
, the value of
d
y
d
x
at
x
=
e
is
Report Question
0%
1
0%
0
0%
-1
0%
none of these
Explanation
Given,
y
=
x
1
/
x
log
y
=
1
x
log
x
⇒
x
log
y
=
log
x
x
y
d
y
d
x
+
log
y
=
1
x
⇒
d
y
d
x
=
y
(
1
−
x
log
y
)
x
2
(
d
y
d
x
)
x
=
e
=
e
1
/
e
(
1
−
e
log
e
1
e
)
e
2
=
0
If for all
x
,
y
the function
f
is defined by
f
(
x
)
+
f
(
y
)
+
f
(
x
)
.
f
(
y
)
=
1
and
f
(
x
)
>
0
then
Report Question
0%
f
′
(
x
)
does not exist
0%
f
′
(
x
)
=
0
for all
x
0%
f
′
(
0
)
<
f
′
(
1
)
0%
none of these
Explanation
f
(
x
)
+
f
(
y
)
+
f
(
x
)
.
f
(
y
)
=
1
put
y
=
0
⇒
f
(
x
)
+
f
(
0
)
+
f
(
x
)
.
f
(
0
)
=
1
⇒
f
(
x
)
=
1
−
f
(
0
)
1
+
f
(
0
)
=
C
(constant)
∴
f
′
(
x
)
=
0
∀
x
If
S
n
denotes the sum of
n
terms of a G.P. whose common ratio is
r
, then
(
r
−
1
)
d
S
n
d
r
is equal to
Report Question
0%
(
n
−
1
)
S
n
+
n
S
n
−
1
0%
(
n
−
1
)
S
n
−
n
S
n
−
1
0%
(
n
−
1
)
S
n
0%
None of these
Explanation
We have,
S
n
=
a
(
r
n
−
1
)
r
−
1
⇒
(
r
−
1
)
S
n
=
a
r
n
−
a
Differentiating both sides with respect to
r
,
we get
(
r
−
1
)
d
S
n
d
r
+
S
n
=
n
a
r
n
−
1
−
0
⇒
(
r
−
1
)
d
S
n
d
r
=
n
a
r
n
−
1
−
S
n
=
n
[
n
th term of
g
.
p
.
]
−
S
n
=
n
(
S
n
−
S
n
−
1
)
−
S
n
=
(
n
−
1
)
S
n
−
n
S
n
−
1
.
If
x
y
=
e
x
+
y
then
d
y
d
x
at
x
=
1
is equal to
Report Question
0%
0
0%
−
2
0%
1
0%
none of these
Explanation
Given,
x
y
=
e
x
+
y
Taking
log
on both sides, we get
y
log
x
=
x
+
y
Taking derivative,
y
x
+
l
o
g
x
.
y
′
=
1
+
y
′
At,
x
=
1
=>
y
=
−
1
So,
y
′
=
−
2
Let
f
(
x
1
+
x
2
+
.
.
.
+
x
n
n
)
=
f
(
x
1
)
+
f
(
x
2
)
+
.
.
.
+
f
(
x
n
)
n
where all
x
i
∈
R
are independent to each other and
n
∈
N
. if
f
(
x
)
is differentiable and
f
′
(
0
)
=
a
,
f
(
0
)
=
b
and
f
′
(
x
)
is equal to
Report Question
0%
a
0%
0
0%
b
0%
None of these
Explanation
Differentiating the given equation w.r.t.
x
1
,
we get
1
n
f
′
(
x
1
+
x
2
+
.
.
.
+
x
n
n
)
=
f
′
(
x
1
)
n
[
Since all
x
′
i
s
are independent to each other,
∴
d
x
i
d
x
j
=
0
if
i
≠
j
and
d
x
i
d
x
j
=
1
if
(
i
=
j
)
]
On putting
x
1
=
x
2
=
.
.
.
=
x
n
−
1
=
0
and
x
n
=
x
,
we get
f
′
(
x
n
)
=
f
′
(
0
)
=
a
.
On integrating, we get
n
f
′
(
x
n
)
=
a
x
+
c
Since
f
(
0
)
=
b
,
we have
c
=
n
b
∴
n
f
′
(
x
n
)
=
a
x
+
n
b
⇒
n
f
′
(
x
)
=
n
a
x
+
n
b
⇒
f
(
x
)
=
a
x
+
b
.
∴
f
′
(
x
)
=
a
,
∀
x
∈
R
If
5
f
(
x
)
+
3
f
(
1
x
)
=
x
+
2
and
y
=
x
f
(
x
)
then
(
d
y
d
x
)
x
=
1
is equal to ?
Report Question
0%
14
0%
7
8
0%
1
0%
none of these
Explanation
5
f
(
x
)
+
3
f
(
1
x
)
=
x
+
2
⇒
(
1
)
Replace
x
by
1
x
⇒
5
f
(
1
x
)
+
3
f
(
x
)
=
1
x
+
2
⇒
(
2
)
⇒
5
×
(
1
)
−
3
×
(
2
)
⇒
16
f
(
x
)
=
5
x
−
3
x
+
4
=
5
x
2
+
4
x
−
3
x
∴
y
=
x
f
(
x
)
=
5
x
2
+
4
x
−
3
16
⇒
d
y
d
x
=
10
x
+
4
16
∴
(
d
y
d
x
)
x
=
1
=
10
+
4
16
=
7
8
y
=
√
sin
x
+
√
sin
x
+
√
sin
x
+
−
∞
then
d
y
d
x
equals:
(
sin
x
>
0
)
Report Question
0%
cos
x
2
y
−
1
0%
y
2
tan
x
+
y
sec
x
0%
1
√
1
+
4
sin
x
0%
2
cos
x
sin
x
+
2
y
Explanation
y
=
√
sin
x
√
sin
x
√
sin
x
+
−
∞
Taking square on both the sides, we get
y
2
=
sin
x
+
√
sin
x
√
sin
x
√
sin
x
+
−
∞
y
2
=
sin
x
+
y
............ Since
y
=
√
sin
x
√
sin
x
√
sin
x
+
−
∞
2
y
d
y
d
x
=
cos
x
+
d
y
d
x
d
y
d
x
=
cos
x
2
y
−
1
If
x
e
x
y
−
y
=
sin
2
x
then
d
y
d
x
at
x
=
0
is
Report Question
0%
0
0%
1
0%
−
1
0%
None of these
Explanation
Given
x
e
x
y
=
y
+
sin
2
x
...(1)
on putting
x
=
0
, we get
0.
e
0
=
y
+
0
⇒
y
=
0
on differentiating (1) both sides w.r.t
x
, we get
1.
e
x
y
+
x
.
e
x
y
(
x
.
d
y
d
x
+
y
)
=
d
y
d
x
+
2
sin
x
cos
x
On putting
x
=
0
,
y
=
0
, we get
e
0
+
0
(
0
+
0
)
=
[
d
y
d
x
]
+
2
sin
0
⇒
d
y
d
x
=
1
If
x
y
.
y
x
=
16
then
d
y
d
x
at (2, 2) is
Report Question
0%
1
0%
−
1
0%
0
0%
none of these
Explanation
x
y
.
y
x
=
16
log
x
y
+
log
y
x
=
log
16
y
log
x
+
x
log
y
=
log
16
Differentiating w.r.t x , we get
y
x
+
log
x
d
y
d
x
+
x
y
d
y
d
x
+
log
y
=
0
So, at x=2, y=2
1
+
log
2
(
d
y
d
x
)
(
2
,
2
)
+
1
(
d
y
d
x
)
(
2
,
2
)
+
log
2
=
0
⇒
(
d
y
d
x
)
(
2
,
2
)
=
−
1
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
1
Not Answered
29
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page