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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 10
l
i
m
n
→
∞
Σ
n
r
=
1
π
n
s
i
n
(
π
r
n
)
is equal to
Report Question
0%
1
0%
2
0%
3
0%
4
Evaluate :
lim
x
→
0
(
e
x
ℓ
n
(
3
x
−
1
)
−
(
3
x
−
1
)
x
sin
x
e
x
ℓ
n
x
)
is equal to
Report Question
0%
1
e
ℓ
n
3
0%
e
ℓ
n
3
0%
3
0%
1
3
L
t
x
→
0
s
i
n
x
−
x
+
x
3
/
6
x
5
=
Report Question
0%
1
120
0%
1
110
0%
1
100
0%
none of these
lim
x
→
0
(
cos
α
)
x
−
(
sin
α
)
x
−
cos
2
α
(
x
−
4
)
,
α
∈
(
0
,
π
2
)
is equal to
Report Question
0%
cos
4
α
.
log
(
cos
α
)
−
sin
4
α
.
log
(
sin
α
)
0%
sin
4
α
.
log
(
cos
α
)
−
cos
4
α
.
log
(
sin
α
)
0%
sin
4
α
.
log
(
cos
α
)
+
cos
4
α
.
log
(
sin
α
)
0%
N
o
n
e
o
f
t
h
e
a
b
o
v
e
lim
x
→
0
3
sin
(
x
9
)
−
sin
(
x
9
)
x
3
=
q
Report Question
0%
0
0%
4
(
π
200
)
3
0%
π
200
0%
π
100
L
t
x
→
0
t
a
n
x
−
x
x
2
t
a
n
x
equals:
Report Question
0%
1
0%
1/2
0%
1/3
0%
None of these
evaluate
l
i
m
x
→
0
x
−
∫
x
0
c
o
s
t
2
d
t
x
3
−
6
x
Report Question
0%
3
0%
−
1
0%
0
0%
1
lim
x
→
0
{
tan
(
π
4
+
x
)
}
1
x
=
Report Question
0%
e
0%
e
2
0%
e
−
1
0%
e
−
2
lim
x
→
∞
(
x
+
1
2
x
+
1
)
x
2
equals?
Report Question
0%
0
0%
e
0%
1
0%
∞
If
g
(
x
)
=
x
[
x
]
f
o
r
x
>
2
t
h
e
n
L
i
m
x
→
2
+
g
(
x
)
−
g
(
2
)
x
−
2
Report Question
0%
-1
0%
0
0%
1/2
0%
1
l
i
m
x
→
∞
2
tan
−
1
x
π
equals
e
L
then
L
is equal to
Report Question
0%
2
π
0%
−
2
π
0%
−
π
2
0%
1
lim
x
→
x
/
2
[
1
−
tan
(
x
2
)
]
[
1
−
sin
x
]
[
1
+
tan
(
x
2
)
]
[
π
−
2
x
]
3
is
Report Question
0%
1
8
0%
0
0%
1
32
0%
∞
lim
θ
→
0
4
θ
(
tan
θ
−
2
θ
tan
θ
)
(
1
−
cos
2
θ
)
2
is
Report Question
0%
1
/
√
2
0%
1
/
2
0%
1
0%
2
the value of
l
i
m
x
→
∞
x
5
5
x
is-
Report Question
0%
0
0%
1
0%
e
5
0%
e
−
5
If
k
is an integer such that
lim
n
→
∞
[
(
cos
k
π
4
)
2
−
(
cos
k
π
6
)
2
]
=
0
then :
Report Question
0%
k
is divisible neither by
4
nor by
8
.
0%
k
must be divisible by
12
but not necessarily by
24
.
0%
k
must be divisible by
24
.
0%
either
k
is divisible by
24
or
k
is neither divisible by
4
nor by
6
.
lim
x
→
5
sin
2
(
x
−
5
)
tan
(
x
−
5
)
(
x
2
−
25
)
(
x
−
5
)
=
Report Question
0%
1
/
10
0%
1
0%
0
0%
−
6
The value of
l
i
m
x
→
0
cos
e
c
4
x
x
2
∫
0
I
n
(
1
+
4
t
)
1
+
t
2
d
t
is
Report Question
0%
1
0%
2
0%
3
0%
4
lim
n
→
∞
n
2
(
x
1
/
n
−
x
1
/
(
n
+
n
)
)
,
x
>
0
, is equal to
Report Question
0%
0
0%
e
x
0%
log
e
x
0%
n
o
n
e
o
f
t
h
e
s
e
lim
x
→
0
l
n
(
s
i
n
3
x
)
l
n
(
s
i
n
x
)
is equal to
Report Question
0%
0
0%
1
0%
2
0%
Non existent
l
i
m
x
→
n
/
2
cot
x
−
cos
x
(
π
−
2
x
)
3
equals
Report Question
0%
1
24
0%
1
16
0%
1
8
0%
1
4
l
i
m
x
→
2
3
√
60
+
x
2
−
4
sin
(
x
−
2
)
equals
Report Question
0%
1
4
0%
0
0%
1
12
0%
Does not exist
lim
x
→
0
sin
x
sin
(
π
3
+
x
)
sin
(
π
3
−
x
)
x
=
Report Question
0%
3
4
0%
1
4
0%
4
3
0%
0
lim
x
→
0
2
(
√
3
sin
(
π
6
+
x
)
−
cos
(
π
6
+
x
)
)
x
√
3
(
√
3
cos
x
−
sin
x
)
Report Question
0%
−
1
/
3
0%
2
/
3
0%
4
/
3
0%
−
4
/
3
lim
x
→
π
/
4
2
√
2
−
(
cos
x
+
sin
x
)
2
1
−
sin
2
x
is equal to
Report Question
0%
1
0%
2
√
2
0%
4
√
2
3
0%
d
o
e
s
n
o
t
e
x
i
s
t
Let
f
be a differentiable function such that
f
′
(
x
)
=
7
−
3
4
f
(
x
)
x
,
(
x
>
0
)
and
f
(
1
)
≠
4
.
Then
lim
x
→
0
+
x
f
(
1
x
)
:
Report Question
0%
Exists and equals 4
0%
Does not exist
0%
Exist and equals
0%
Exists and equals
4
7
Explanation
f
′
(
x
)
=
7
−
3
4
f
(
x
)
x
(
x
>
0
)
Given
f
(
1
)
≠
4
lim
x
→
0
+
x
f
(
1
x
)
=
?
d
y
d
x
+
3
4
y
x
=
7
(This is LDE)
IF
=
e
∫
3
4
x
d
x
=
e
3
4
l
n
|
x
|
=
x
3
4
y
.
x
3
4
=
∫
7.
x
3
4
d
x
y
.
x
3
4
=
7.
x
7
4
7
4
+
C
f
(
x
)
=
4
x
+
C
.
x
−
3
4
f
(
1
x
)
=
4
x
+
C
.
x
3
4
lim
x
→
0
+
x
f
(
1
x
)
=
lim
x
→
0
+
(
4
+
C
.
x
7
4
)
=
4
If
L
i
m
x
→
0
(
1
+
a
3
)
+
8
e
1
/
x
1
+
(
1
−
b
3
)
+
e
1
/
x
=
2
then
Report Question
0%
a
=
1
,
b
=
2
0%
a
=
1
,
b
=
−
3
1
/
3
0%
a
=
2
,
b
=
3
1
/
3
0%
none of these
l
i
m
x
→
π
2
t
a
n
2
x
(
√
2
s
i
n
2
x
+
3
s
i
n
x
+
4
−
√
s
i
n
2
x
+
6
s
i
n
x
+
2
)
is equal to
Report Question
0%
3
4
0%
1
6
0%
1
12
0%
5
12
lim
x
→
0
1
−
cos
x
x
log
(
1
+
x
)
=
Report Question
0%
1
0%
0
0%
−
1
0%
1
2
Evaluate
lim
x
→
0
sin
[
cos
x
]
1
+
[
cos
x
]
(
[
.
]
denotes the greatest integer function)
Report Question
0%
−
1
0%
1
0%
0
0%
d
o
e
s
n
o
t
e
x
i
s
t
lim
x
→
0
x
(
e
sin
x
−
1
)
1
−
cos
x
Report Question
0%
1
2
0%
2
0%
0
0%
1
P
=
lim
x
→
o
+
(
1
+
tan
2
√
x
)
1
/
2
x
=
____
Report Question
0%
P
=
1
/
4
0%
P
=
1
/
2
0%
P
=
2
0%
P
=
1
lim
x
→
0
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
n
1
/
x
n
)
n
x
,
n
ϵ
N
is equal to
Report Question
0%
n
!
0%
1
0%
1
n
!
0%
0
Explanation
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
.
.
.
.
.
.
+
n
1
/
x
n
)
n
x
if we put the limits then we have
1
∞
form. We have for
l
i
m
x
→
∞
[
f
(
x
)
]
9
(
x
)
such that limit has the form
1
∞
, then
l
i
m
x
→
∞
[
f
(
x
)
]
9
(
x
)
=
e
l
i
m
x
→
∞
9
(
x
)
[
f
(
x
)
−
1
]
⟶
(
1
)
Using this we can evalute the limit asked,
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
.
n
1
/
x
n
)
n
x
=
e
l
i
m
x
→
∞
n
x
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
.
.
.
.
.
.
n
1
/
x
−
n
n
)
=
e
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
.
n
1
/
x
−
n
(
1
/
x
)
)
if we put the limits we find that it is in
÷
form, so we will employ the L'Hospital's rule
=
e
l
i
m
x
→
∞
(
1
1
/
x
(
−
1
x
2
)
l
n
1
+
2
1
/
x
(
−
1
x
2
)
l
n
2......
+
n
1
/
x
(
−
1
x
2
)
l
n
n
−
0
(
−
1
/
x
2
)
)
=
e
l
i
m
x
→
∞
(
1
1
/
x
l
n
1
+
2
1
/
x
l
n
2......
+
n
1
/
x
l
n
n
)
=
e
l
n
1
+
l
n
2
+
l
n
3......
+
l
n
n
[
∵
l
i
m
x
→
∞
n
1
/
x
=
n
0
=
1
]
=
e
l
n
(
1.2.3.4.5.6......
n
)
[
∵
l
n
a
+
l
n
b
=
l
n
a
b
]
=
e
l
n
n
!
=
n
!
[
∵
e
l
n
b
=
b
]
Answer : Option A
The values of
lim
n
→
∞
4
√
n
5
+
2
−
3
√
n
2
+
1
5
√
n
4
+
2
−
2
√
n
3
+
1
is?
Report Question
0%
1
0%
0
0%
−
1
0%
∞
lim
x
→
0
x
tan
2
x
−
2
x
tan
x
(
1
−
cos
2
x
)
2
equal to
Report Question
0%
1
4
0%
1
0%
1
2
0%
N
o
n
e
o
f
t
h
e
s
e
Let
f
(
x
)
=
lim
n
→
∞
2
x
2
n
sin
1
x
+
x
1
+
x
2
x
then find
Report Question
0%
lim
x
→
∞
x
f
(
x
)
0%
lim
x
→
1
f
(
x
)
0%
lim
x
→
0
f
(
x
)
0%
lim
x
→
−
∞
f
(
x
)
If
lim
x
→
0
(
cos
x
+
a
3
sin
(
b
6
x
)
)
1
x
=
e
512
then value of
a
b
2
is equal to
Report Question
0%
−
512
0%
512
0%
8
0%
none of these
i
f
(
x
)
=
[
x
−
3
]
+
[
x
−
4
]
f
o
r
x
ϵ
R
t
h
e
n
l
i
m
x
→
3
f
(
x
)
=
Report Question
0%
-2
0%
-1
0%
0
0%
2
lim
x
→
0
(
cos
x
)
1
/
2
−
(
cos
x
)
1
/
3
sin
2
x
is $$
Report Question
0%
1
/
6
0%
−
1
/
12
0%
2
/
3
0%
1
/
3
The value of
l
i
m
x
→
y
s
i
n
2
x
−
s
i
n
2
y
x
2
−
y
2
Report Question
0%
0
0%
1
0%
s
i
n
y
y
0%
s
i
n
2
y
2
y
lim
x
→
0
x
tan
2
x
−
2
x
tan
x
(
1
−
cos
2
x
)
2
is equal to?
Report Question
0%
2
0%
−
2
0%
1
/
2
0%
−
1
/
2
l
i
m
x
→
0
(
cos
x
+
a
sin
b
x
)
1
x
is equal to
Report Question
0%
e
a
0%
e
a
b
0%
e
b
0%
e
a
/
b
lim
x
→
0
x
(
1
+
a
cos
x
)
−
b
sin
x
x
3
=
1
then
Report Question
0%
a
=
−
5
/
2
0%
a
=
−
3
/
2
,
b
=
−
1
/
2
0%
a
=
−
3
/
2
,
b
=
−
5
/
2
0%
a
=
−
5
/
2
,
b
=
−
3
/
2
If
lim
x
→
0
[
1
+
x
l
n
(
1
+
b
2
)
]
1
/
x
=
2
b
sin
2
θ
,
b
>
s m f
θ
∈
(
−
π
,
π
]
, then the value of
θ
is?
Report Question
0%
±
π
4
0%
±
π
3
0%
±
π
6
0%
±
π
2
lim
x
→
∞
sin
(
π
cos
−
2
x
)
x
2
is equal to:
Report Question
0%
−
π
0%
π
0%
π
2
0%
1
The value of
lim
θ
→
0
+
sin
√
θ
√
sin
θ
is equal to?
Report Question
0%
0
0%
1
0%
−
1
0%
4
Let
f
:
(
π
2
,
π
2
)
→
R
,
f
(
x
)
=
{
lim
n
→
∞
(
t
a
n
x
)
2
n
+
x
2
s
i
n
2
x
+
(
t
a
n
x
)
2
n
;
x
≠
0
1
;
x
=
0
,
n
∈
N
. Which of the following holds good ?
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0%
f
(
−
π
−
4
)
=
f
(
π
+
4
)
0%
f
(
−
π
−
4
)
=
f
(
−
π
+
4
)
0%
f
(
π
−
4
)
=
f
(
π
+
4
)
0%
f
(
0
+
)
=
f
(
0
)
=
f
(
0
)
lim
x
→
0
(
1
sin
2
x
−
1
sin
h
2
x
)
=
?
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0%
2
/
3
0%
0
0%
1
/
3
0%
−
2
/
3
Consider
lim
x
→
0
a
x
+
b
e
−
x
+
sin
x
+
1
a
x
−
b
sin
x
=
ℓ
(
ℓ
is a finite number)
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0%
ℓ
=
0
,
if
a
=
−
2
,
b
=
−
1
0%
ℓ
=
5
,
i
f
a
=
1
,
b
=
−
1
0%
ℓ
=
1
,
if
a
=
b
=
−
1
0%
None of these
lim
x
→
0
sin
(
π
cos
2
x
)
x
2
is equal to?
Report Question
0%
−
π
0%
π
0%
π
/
2
0%
1
If
f
(
x
)
is odd linear polynomial with
f
(
1
)
=
1
, then
lim
x
→
0
2
f
(
tan
x
)
−
2
f
(
sin
x
)
x
2
f
(
sin
x
)
is
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0%
1
0%
ln
2
0%
1
2
ln
2
0%
cos
2
0:0:1
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0
Answered
1
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Incorrect : 0
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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