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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 12
lim
x
→
∞
sin
x
sin
(
π
3
+
x
)
sin
(
π
3
−
x
)
x
=
Report Question
0%
3
4
0%
1
4
0%
4
3
0%
0
If
α
,
β
,
i
n
(
−
π
2
,
0
)
such that
(
s
i
n
α
+
s
i
n
β
)
+
s
i
n
α
s
i
n
β
=
0
and
(
s
i
n
α
+
s
i
n
β
)
s
i
n
α
s
i
n
β
=
−
1
and
λ
=
l
i
m
n
→
∞
1
+
(
2
s
i
n
t
h
e
t
a
)
2
n
(
2
s
i
n
b
e
t
a
)
2
n
then
Report Question
0%
2
α
+
3
β
=
−
5
π
6
0%
λ
π
+
α
+
β
=
5
π
6
0%
α
−
β
=
π
3
0%
α
+
β
=
−
π
3
lim
x
→
0
27
x
−
9
x
−
3
x
+
1
√
2
−
√
1
+
cos
x
=
Report Question
0%
0
0%
8
√
2
(
log
3
)
2
0%
8
(
log
3
)
2
0%
1
lim
x
→
0
{
(
s
i
n
x
−
x
)
/
x
3
)
}
equals:
Report Question
0%
1
/
3
0%
−
1
/
3
0%
1
/
6
0%
−
1
/
6
lim
x
→
0
1
x
8
[
1
−
cos
(
x
2
2
)
]
[
1
−
cos
(
x
2
4
)
]
Report Question
0%
1
8
0%
1
8
2
0%
1
8
3
0%
1
8
4
lim
x
→
a
(
2
−
a
x
)
tan
(
π
x
2
a
)
Report Question
0%
e
−
a
π
0%
e
−
2
a
π
0%
e
−
2
x
0%
1
Solve
L
i
m
x
→
0
8
x
8
(
1
−
c
o
s
x
2
2
−
c
o
s
x
2
4
+
c
o
s
x
2
2
.
c
o
s
x
2
4
)
=
Report Question
0%
1
16
0%
1
15
0%
1
32
0%
1
If
f
(
x
)
=
√
1
−
√
1
−
x
2
, then
f
(
x
)
is
Report Question
0%
continuous on
[
−
1
,
1
]
0%
differentiable on
(
−
1
,
0
)
∪
(
0
,
1
)
0%
both (a) and (b)
0%
None of the above
l
i
m
x
→
0
8
x
8
(
1
−
c
o
s
x
2
2
−
c
o
s
x
2
4
+
c
o
s
x
2
2
.
c
o
s
x
2
4
)
=
Report Question
0%
1
16
0%
1
15
0%
1
32
0%
1
lim
x
→
0
[
100
tan
x
.
sin
x
x
2
]
where
[
.
]
represents greatest integer function is
Report Question
0%
99
0%
100
0%
0
0%
98
The value of
lim
x
→
1
2
2
sin
−
1
x
−
π
2
1
−
2
x
2
is equal to
Report Question
0%
−
1
0%
0
0%
1
0%
2
f
(
x
)
=
log
1
−
2
x
(
1
+
2
x
)
for
x
≠
0
=
k
for
x
=
0
is continuous at
x
=
0
, find
k
.
Report Question
0%
1
0%
−
1
0%
0
0%
1
2
The value of
lim
x
→
π
2
tan
2
(
√
2
sin
2
x
+
3
sin
x
+
4
−
√
sin
2
x
+
6
sin
x
+
2
)
is equal to
Report Question
0%
0
0%
1
11
0%
1
12
0%
1
8
lim
x
→
∞
(
x
2
sin
(
1
x
)
−
x
1
−
|
x
|
)
=
Report Question
0%
0
0%
1
0%
-1`
0%
2
lim
x
→
0
(
e
x
+
e
−
x
−
2
x
2
)
1
/
x
2
Report Question
0%
e
1
/
2
0%
e
1
/
4
0%
e
1
/
3
0%
e
1
/
12
The value of
lim
x
→
0
log
cos
2
x
cos
x
+
log
cos
2
x
cos
2
x
equals
Report Question
0%
5
4
0%
17
4
0%
13
16
0%
29
10
l
i
m
x
→
0
(
c
o
s
+
s
i
n
x
)
1
/
x
is equal to
Report Question
0%
e
0%
e
2
0%
e
−
1
0%
1
lim
x
→
0
sin
[
cos
x
]
1
+
[
cos
x
]
is (where [] is G.I.F)
Report Question
0%
1
0%
0
0%
does not exist
0%
2
l
i
m
x
→
a
(
sin
x
−
a
2
tan
π
x
2
a
)
Report Question
0%
a
/
π
0%
−
a
/
π
0%
π
/
a
0%
−
π
/
a
Let
a
∈
(
0
,
π
2
)
, then the value of
lim
a
→
0
1
a
3
∫
a
0
ℓ
n
(
1
+
t
a
n
a
t
a
n
x
)
d
x
is equal to
Report Question
0%
1
3
0%
1
2
0%
1
6
0%
1
lim
x
→
0
sin
x
x
=
y
Report Question
0%
y
>
1
0%
y
<
1
0%
y
≥
1
0%
y
≤
1
The value of
Limit
x
→
0
cos
(
sin
x
)
−
cos
x
x
4
is equal to
Report Question
0%
1/5
0%
1/6
0%
1/4
0%
1/2
lim
x
→
0
1
−
c
o
s
3
x
x
s
i
b
x
c
o
s
x
is equal to
Report Question
0%
2/5
0%
3/5
0%
3/2
0%
3/4
The value of
lim
x
→
0
sin
3
(
√
x
)
ln
(
1
+
3
x
)
(
tan
−
1
√
x
)
2
(
e
5
(
√
x
)
−
1
)
x
is equal to
Report Question
0%
1
5
0%
3
5
0%
2
5
0%
4
5
L
i
m
x
→
0
s
i
n
x
x
=
y
Report Question
0%
y
>
1
0%
y
<
1
0%
y
≥
1
0%
y
≤
1
L
t
x
→
1
(
1
−
x
)
tan
π
x
=
Report Question
0%
−
1
0%
0
0%
1
0%
2
l
i
m
x
→
a
(
2
−
a
x
)
t
a
n
(
π
x
2
a
)
is equal to
Report Question
0%
e
−
a
π
0%
e
−
2
a
π
0%
e
−
2
π
0%
1
lim
x
→
0
(
e
x
+
e
−
x
−
2
x
2
)
1
/
x
2
is
Report Question
0%
e
1
/
2
0%
e
1
/
4
0%
e
1
/
3
0%
e
1
/
12
Let x be an irrational, then
l
i
m
m
→
∞
l
i
m
n
→
∞
{
c
o
s
(
n
!
π
x
)
}
2
m
equals
Report Question
0%
0
0%
-1
0%
1
0%
Indeterminate
L
i
m
→
1
−
c
o
s
2
x
x
s
i
n
2
x
Report Question
0%
1/2
0%
3/2
0%
3/4
0%
1/4
x
l
i
m
→
5
(
√
1
−
cos
(
2
x
−
10
)
sin
(
x
−
5
)
)
Report Question
0%
−
√
2
0%
√
2
0%
does not exist
0%
none of these
The value of
lim
x
→
0
sec
5
x
−
sec
3
x
sec
3
x
−
sec
x
Report Question
0%
-2
0%
1
0%
2
0%
-1/2
lim
x
→
0
∫
x
0
(
tan
−
1
t
)
2
√
1
+
x
2
d
t
is equal to
Report Question
0%
π
2
0%
π
2
2
0%
π
2
4
0%
None of these
the value of
l
i
m
x
⟶
∞
X
4
s
i
n
(
1
x
)
+
x
3
1
+
|
x
|
3
Report Question
0%
1
0%
-1
0%
2
0%
does not exist
l
i
m
x
→
1
[
c
o
s
e
c
π
x
2
]
1
/
(
1
−
x
)
(where
[
.
]
represents the greatest integer function) is equal to
Report Question
0%
0
0%
1
0%
∞
0%
D
o
e
s
n
o
t
e
x
i
s
t
f(X)=|x|+|x-1| is continuous at
Report Question
0%
'0' only
0%
0,1 only
0%
Every where
0%
No where
Explanation
Step-1: Finding values of
f
(
x
)
f
(
x
)
=
|
x
|
+
|
x
−
1
|
f
(
x
)
=
−
x
−
|
x
−
1
|
at
x
≤
0
=
x
−
(
x
−
1
)
at
0
≤
x
<
1
=
x
+
(
x
−
1
)
at
x
≥
1
Hence, we get
f
(
x
)
=
1
−
2
x
at
x
≤
0
=
1
at
0
≤
x
<
1
=
2
x
−
1
at
x
≥
1
Step-2: Finding limits of the function at constraint values
At
x
=
0
,
lim
x
→
0
f
(
x
)
=
1
Hence,
f
(
x
)
is continuous at
x
=
0
At
x
=
1
,
lim
x
→
1
f
(
x
)
=
1
Hence,
f
(
x
)
is continuous at
x
=
1
Hence,
f
(
x
)
is continuous everywhere
Hence,Correct option is (C)
Find:
l
i
m
x
→
0
1
−
c
o
s
3
x
x
s
i
n
2
x
=
Report Question
0%
1/2
0%
3/2
0%
3/4
0%
1/4
lim
x
→
0
sin
x
−
x
x
3
is equal to
Report Question
0%
−
1
6
0%
−
1
3
0%
−
1
2
0%
−
1
lim
x
→
0
1
−
cos
x
x
log
(
1
+
x
)
=
Report Question
0%
1
0%
0
0%
-1
0%
1/2
If
α
and
β
be the roots of the equation
a
x
2
+
b
x
+
c
=
0
then
lim
x
→
1
α
√
1
−
cos
2
(
c
x
2
+
b
x
+
a
)
4
(
1
−
α
x
)
2
Report Question
0%
Does not exist
0%
Equals
|
c
2
α
(
1
α
+
1
β
)
|
0%
Equals
|
c
2
α
(
1
α
−
1
β
)
|
0%
Equals
|
c
2
(
1
α
+
1
β
)
|
The value of
l
i
m
x
→
1
√
2
x
−
c
o
s
(
s
i
n
−
1
x
)
1
−
t
a
n
(
s
i
n
−
1
x
)
i
s
Report Question
0%
−
1
√
2
0%
1
√
2
0%
√
2
0%
−
√
2
l
i
m
x
→
π
/
2
sin
(
x
c
o
s
x
)
c
o
s
(
x
s
i
n
x
)
is equal to
Report Question
0%
1
0%
π
2
0%
2
π
0%
does not exist
l
i
m
x
→
0
x
c
o
t
(
4
x
)
s
i
n
2
x
c
o
t
2
(
2
x
)
is equal to
Report Question
0%
0
0%
2
0%
1
0%
4
l
i
m
x
→
0
s
e
c
4
x
−
s
e
c
2
x
s
e
c
3
x
−
s
e
c
x
=
Report Question
0%
3/2
0%
2/3
0%
1/3
0%
3/4
The value of
l
i
m
x
→
1
(
2
−
x
)
t
a
n
(
π
x
2
)
is
Report Question
0%
e
−
2
/
p
i
0%
e
1
/
p
i
0%
e
2
/
p
i
0%
e
−
1
/
p
i
l
i
m
x
→
1
x
2
−
1
sin
2
x
+
cos
x
cos
(
x
+
2
)
−
cos
2
(
x
+
1
)
is-
Report Question
0%
0
0%
1
cos
1
0%
2
sin
2
0%
1
2
cos
1
lim
x
→
1
1
−
x
−
2
/
3
1
−
x
−
1
/
3
Report Question
0%
2
0%
1
0%
0
0%
does not exist
l
i
m
n
→
∞
−
3
n
+
(
−
1
)
n
4
n
−
(
−
1
)
n
is equal to
Report Question
0%
$$-\dfrac{3]{4}$$
0%
o if n is even
0%
−
3
4
if n is odd
0%
None of these
lim
x
→
1
(
x
4
+
x
2
+
x
+
1
x
2
−
x
+
1
)
1
−
cos
(
x
+
1
)
(
x
+
1
)
2
is equal to:
Report Question
0%
1
0%
(
2
/
3
)
1
/
2
0%
(
3
/
2
)
1
/
2
0%
e
1
/
2
lim
x
→
0
(
1
x
+
2
x
+
3
x
+
⋯
+
n
x
n
)
1
/
x
is equal to
Report Question
0%
(
n
!
)
n
0%
(
n
!
)
1
/
n
0%
n
!
0%
ln
(
n
!
)
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