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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 1 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 1
Evaluate
lim
x
→
3
4
√
x
3
using the properties of limits.
Report Question
0%
28
1
/
4
0%
25
1
/
4
0%
27
1
/
4
0%
26
1
/
4
Explanation
lim
x
→
3
4
√
x
3
=
(
3
3
)
1
/
4
=
(
27
)
1
/
4
I
f
A
i
=
x
−
a
i
|
x
−
a
i
|
,
i
=
1
,
2
,
3
,
.
.
.
.
.
n
and
a
1
<
a
2
<
a
3
.
.
.
.
<
a
n
,
t
h
e
n
lim
x
→
a
m
(
A
1
A
2
.
.
.
.
.
.
A
n
)
,
1
≤
m
≤
n
Report Question
0%
is equal to
(
−
1
)
m
0%
is equal to
(
−
1
)
m
+
1
0%
is equal to
(
−
1
)
n
−
m
0%
is equal to
(
−
1
)
n
−
m
−
1
lim
x
→
π
2
cot
x
−
cos
x
(
π
2
−
x
)
3
Report Question
0%
−
1
2
0%
1
2
0%
2
0%
1
f
(
x
)
=
x
sin
1
x
,
f
o
r
x
≠
0
=
0
,
f
o
r
x
=
0
Then.
Report Question
0%
f
′
(
0
+
)
e
x
i
t
b
u
t
f
′
(
0
−
)
does not exit
0%
f
′
(
0
+
)
a
n
d
f
′
(
0
−
)
do not exit
0%
f
′
(
0
+
)
=
f
′
(
0
−
)
0%
none of these
lim
x
→
0
sin
−
1
x
−
tan
−
1
x
x
3
is equal to
Report Question
0%
0
0%
1
0%
−
1
0%
1
2
If
lim
x
→
∞
(
x
2
+
x
+
1
x
+
1
−
a
x
−
b
)
=
4
,then
Report Question
0%
a
=
1
,
b
=
4
0%
a
=
1
,
b
=
−
4
0%
a
=
2
,
b
=
−
3
0%
a
=
2
,
b
=
3
lim
n
→
∞
(
tan
θ
+
1
2
tan
θ
2
+
1
2
2
tan
θ
2
2
+
.
.
.
+
1
2
n
tan
θ
2
n
)
equals?
Report Question
0%
1
θ
0%
1
θ
−
2
cot
2
θ
0%
2
cot
2
θ
0%
None of these
Let p=
lim
x
→
0
+
(
1
+
t
a
n
2
√
x
)
1
2
x
then log p is equal to :
Report Question
0%
1
0%
1
2
0%
1
4
0%
2
lim
x
→
π
4
(
sin
2
x
)
sec
2
2
x
is equal to
Report Question
0%
−
1
2
0%
1
2
0%
e
−
1
2
0%
e
1
2
lim
θ
→
π
/
2
1
−
sin
θ
(
π
/
2
−
θ
)
cos
θ
is equal to
Report Question
0%
1
0%
−
1
0%
1
/
2
0%
−
1
/
2
If
l
i
m
x
→
∞
(
1
+
a
x
−
4
x
2
)
2
x
=
e
3
, then 'a' is equal to :
Report Question
0%
2
0%
3
2
0%
2
3
0%
1
2
The value of
l
i
m
x
→
0
s
i
n
(
π
c
o
s
2
x
)
x
2
equals
Report Question
0%
−
π
0%
π
0%
π
2
0%
2
π
l
i
m
x
→
1
x
t
a
n
(
x
−
[
x
]
)
x
−
1
is:
Report Question
0%
1
0%
0
0%
-1
0%
does not exist
The value of
l
i
m
θ
→
π
2
(
s
e
c
θ
−
t
a
n
θ
)
equals
Report Question
0%
0
0%
1
0%
2
0%
∞
The value of
lim
x
→
0
1
−
cos
3
x
x
sin
x
cos
x
is
Report Question
0%
2
5
0%
3
5
0%
3
2
0%
3
4
The value of
lim
x
→
∞
a
x
sin
(
b
a
x
)
where
a
>
1
is
Report Question
0%
b
log
a
0%
a
log
b
0%
b
0%
a
The value of
lim
x
→
3
(
x
3
+
27
)
log
e
(
x
−
2
)
x
2
−
9
is
Report Question
0%
9
0%
18
0%
27
0%
1
3
The value of
lim
x
→
x
4
4
√
2
−
(
cos
x
+
sin
x
)
5
1
−
sin
2
x
is
Report Question
0%
0
0%
√
2
0%
5
√
2
0%
3
lim
x
→
0
(
[
−
5
sin
x
x
]
+
[
6
sin
x
x
]
)
(where
[
.
]
denotes greatest integer function) is equal to
Report Question
0%
0
0%
−
12
0%
1
0%
2
If
l
i
m
x
→
0
x
3
√
a
+
x
(
b
x
−
s
i
n
x
)
=
1
,
a > 0, then a + b is equal to
Report Question
0%
36
0%
37
0%
38
0%
40
lim
x
→
0
sin
x
5
sin
4
x
=
Report Question
0%
0
0%
1
0%
−
1
0%
∞
lim
x
→
0
(
1
−
cos
2
x
)
sin
5
x
x
2
sin
3
x
=
?
Report Question
0%
10
/
3
0%
3
/
10
0%
6
/
5
0%
56
l
i
m
x
→
0
1
−
cos
2
x
cos
2
x
−
cos
8
x
is equal to
Report Question
0%
−
1
/
15
0%
1
/
10
0%
1
/
15
0%
15
lim
x
→
−
1
cos
2
−
cos
2
x
x
2
−
|
x
|
is equal to :
Report Question
0%
0
0%
cos
2
0%
2
sin
2
0%
None
lim
x
→
0
x
.
10
x
−
x
1
−
c
o
s
x
=
Report Question
0%
log
10
0%
2
log
10
0%
3
log
10
0%
4
log
10
lim
x
→
0
3
√
1
+
sin
x
−
3
√
1
−
sin
x
x
=
Report Question
0%
0
0%
1
0%
2
3
0%
3
2
l
i
m
x
→
0
s
i
n
(
6
x
2
)
I
n
c
o
s
(
2
x
2
−
x
)
=
Report Question
0%
12
0%
-12
0%
6
0%
-6
lim
x
→
1
(
log
3
3
x
)
log
x
3
=
?
Report Question
0%
e
−
1
0%
e
0%
−
1
0%
1
Let
f
(
β
)
=
lim
α
→
β
sin
2
α
−
sin
2
β
α
2
−
β
2
,
then
f
(
π
4
)
is greater than-
Report Question
0%
lim
x
→
0
1
−
cos
3
x
x
sin
2
x
0%
lim
x
→
π
/
2
cot
x
−
cos
x
(
π
−
2
x
)
3
0%
lim
x
→
∞
(
cos
√
x
+
1
−
cos
√
x
)
0%
lim
x
→
a
√
a
+
2
x
−
√
3
x
√
3
a
+
x
−
2
√
x
where
a
>
0
l
i
m
x
→
−
∞
(
3
x
4
+
2
x
2
)
s
i
n
(
1
x
)
+
|
x
|
3
+
5
|
x
|
3
+
|
x
|
2
+
|
x
|
+
1
=
Report Question
0%
2
0%
1
0%
-2
0%
-3
The value f
lim
x
→
π
/
4
√
1
−
√
sin
2
x
π
−
4
x
=
Report Question
0%
−
1
4
0%
1
4
0%
1
2
0%
N
o
n
e
o
f
t
h
e
s
e
If
L
=
l
i
m
x
→
0
a
s
i
n
x
−
s
i
n
2
x
t
a
n
3
x
is finite, then the value of L is :
Report Question
0%
1
0%
2
0%
3
0%
-1
If
lim
x
→
0
a
e
−
x
−
b
cos
x
−
1
2
c
x
x
cos
x
=
2
then the value of
a
+
b
+
c
is-
Report Question
0%
4
0%
−
4
0%
2
0%
−
2
lim
x
→
0
sin
2
x
+
3
x
2
x
+
sin
3
x
is equal to
Report Question
0%
1
0%
1
5
0%
2
0%
Does not exist
If
f
(
x
)
is continuous and
f
(
9
2
)
=
2
9
, then
lim
x
→
0
f
(
1
−
cos
3
x
x
2
)
is equal to :
Report Question
0%
9
2
0%
2
9
0%
0
0%
8
9
Explanation
lim
x
→
0
1
−
c
o
s
3
x
x
2
=
lim
x
→
0
2
sin
2
(
3
x
2
)
(
3
x
2
)
2
×
9
4
=
9
2
⇒
lim
x
→
0
f
(
1
−
c
o
s
3
x
x
2
)
=
f
(
9
2
)
=
2
9
.
The function
f
:
(
R
−
0
)
→
R given by
f
(
x
)
=
1
x
−
2
e
2
x
−
1
can be made continuous at
x
=
0
by defining
f
(
0
)
as
Report Question
0%
2
0%
−
1
0%
0
0%
1
Explanation
Given
f
(
x
)
=
1
x
−
2
e
2
x
−
1
⇒
f
(
0
)
=
lim
x
→
0
{
1
x
−
2
e
2
x
−
1
}
=
lim
x
→
0
e
2
x
−
1
−
2
x
x
(
e
2
x
−
1
)
.
.
.
.
.
.
.
[
0
0
f
o
r
m
]
∴
usingL'Hospital rule
f
(
0
)
=
lim
x
→
0
2
e
2
x
−
2
(
e
2
x
−
1
+
2
x
e
2
x
)
f
(
0
)
=
lim
x
→
0
4
e
2
x
4
x
e
2
x
+
2
e
2
x
+
2
e
2
x
=
4.
e
0
4
(
0
+
e
0
)
=
1
If the function
f
(
x
)
=
{
√
2
+
cos
x
−
1
(
π
−
x
)
2
x
≠
π
k
x
=
π
is continuous at
x
=
π
, then
k
equals :
Report Question
0%
0
0%
1
2
0%
2
0%
1
4
Explanation
lim
x
→
π
√
2
+
cos
x
−
1
(
π
−
x
)
2
=
lim
x
→
π
1
+
cos
x
(
π
−
x
)
2
(
√
2
+
cos
x
+
1
)
[Rationalize numerator and denominator ]
=
lim
x
→
π
1
−
cos
(
π
−
x
)
(
π
−
x
)
2
(
√
2
+
cos
x
+
1
)
=
lim
x
→
π
2
sin
2
(
π
−
x
)
2
(
π
−
x
)
2
(
√
2
+
cos
x
+
1
)
=
1
4
So,
k
=
1
4
.
If
f
:
R
→
R
is a function defined by
f
(
x
)
=
[
x
]
cos
(
2
x
−
1
2
π
)
, where
[
x
]
denotes the greatest integer function, then
f
is:
Report Question
0%
continuous for every real
x
0%
discontinuous only at
x
=
0
0%
discontinuous only at non-zero integral values of
x
0%
continuous only at
x
=
0
Explanation
For
n
∈
I
lim
x
→
n
+
f
(
x
)
=
lim
x
→
n
+
[
x
]
cos
(
2
x
−
1
2
)
π
=
n
cos
(
2
n
−
1
2
)
π
=
0
And
lim
x
→
n
−
f
(
x
)
=
lim
x
→
n
−
[
x
]
cos
(
2
x
−
1
2
)
π
=
(
n
−
1
)
cos
(
2
n
−
1
2
)
π
=
0
Thus
f
is continuous for
x
=
n
∈
I
........ (1)
Since the function
g
(
x
)
=
[
x
]
and
h
(
x
)
=
cos
(
2
x
−
1
2
)
π
are continuous on
x
∈
R
−
I
, ......... (2)
From, (1) and (2), we get
f
is continuous everywhere
lim
x
→
0
x
2
sin
π
x
=
Report Question
0%
1
0%
0
0%
does not exist
0%
∞
Explanation
s
i
n
π
x
behaves\ as\ oscillatory\ function\ at
x
→
0
b
u
t
h
a
v
e
f
i
n
i
t
e
v
a
l
u
e
=
0
0
×
f
i
n
i
t
e
v
a
l
u
e
=
0
lim
x
→
0
x
e
x
−
sin
x
x
is equal to
Report Question
0%
3
0%
1
0%
0
0%
2
Explanation
lim
x
→
0
x
e
x
−
sin
x
x
=
lim
x
→
0
x
e
x
x
−
lim
x
→
0
sin
x
x
, split up
=
lim
x
→
0
e
x
−
lim
x
→
0
sin
x
x
=
e
0
−
1
=
1
−
1
=
0
Use limit properties to evaluate
lim
x
→
4
3
x
2
tan
π
x
x
Report Question
0%
12
0%
14
0%
16
0%
18
Explanation
lim
x
→
4
3
x
2
tan
π
x
x
=
lim
x
→
4
(
3
x
tan
π
x
)
, simplify
=
3
(
4
)
tan
π
4
=
12
(
1
)
=
12
, [substitute the limit ]
The from of the limit is not indeterminate, so we can substitute the limit value
For every integer
n
, let
a
n
and
b
n
be real numbers. Let function
f
:
I
R
→
I
R
be given by
f
(
x
)
=
{
a
n
+
sin
π
x
,
f
o
r
x
∈
[
2
n
,
2
n
+
1
]
b
n
+
cos
π
x
,
f
o
r
x
∈
(
2
n
−
1
,
2
n
)
,
f
o
r
a
l
l
i
n
t
e
g
e
r
s
n
.
lf
f
is continuous, then which of the following hold(s) for all
n
?
Report Question
0%
a
n
−
1
−
b
n
−
1
=
0
0%
a
n
−
b
n
=
1
0%
a
n
−
b
n
+
1
=
1
0%
a
n
−
1
−
b
n
=
−
1
Explanation
Check the continuity at
x
=
2
n
L
.
H
.
L
=
lim
x
→
2
n
−
[
f
(
x
)
]
=
lim
h
→
0
[
f
(
2
n
−
h
)
]
=
lim
h
→
0
[
b
n
+
cos
π
(
2
n
−
h
)
]
=
lim
h
→
0
[
b
n
+
cos
π
h
]
=
b
n
+
1
R
.
H
.
L
=
lim
x
→
2
n
+
[
f
(
x
)
]
=
lim
h
→
0
[
f
(
2
n
+
h
)
]
=
lim
h
→
0
[
a
n
+
sin
π
(
2
n
+
h
)
]
=
a
n
f
(
2
n
)
=
a
n
+
sin
2
π
n
=
a
n
For continuity,
lim
x
→
2
n
−
f
(
x
)
=
lim
x
→
2
n
+
f
(
x
)
=
f
(
2
n
)
So,
a
n
=
b
n
+
1
⇒
a
n
−
b
n
=
1
Now, check for continuity at
x
=
2
n
+
1
L
.
H
.
L
.
=
lim
h
→
0
(
a
n
+
sin
π
(
2
n
+
1
−
h
)
)
=
a
n
R
.
H
.
L
.
=
lim
h
→
0
(
b
n
+
1
+
cos
(
π
(
2
n
+
1
−
h
)
)
)
=
b
n
+
1
−
1
f
(
2
n
+
1
)
=
a
n
.
For continuity,
a
n
=
b
n
+
1
−
1
⇒
a
n
−
1
=
b
n
−
1
Both, option B and option D are correct.
If
lim
x
→
∞
x
sin
(
1
x
)
=
A
and
lim
x
→
0
x
sin
(
1
x
)
=
B
, then which one of the following is correct?
Report Question
0%
A
=
1
and
B
=
0
0%
A
=
0
and
B
=
1
0%
A
=
0
and
B
=
0
0%
A
=
1
and
B
=
1
Explanation
As given
A
=
lim
x
→
∞
x
sin
(
1
x
)
=
lim
x
→
∞
sin
(
1
x
)
(
1
x
)
Let
t
=
1
x
;
x
→
α
,
t
→
0
⇒
A
=
lim
t
→
∞
sin
t
t
=
1
[
∵
lim
t
→
0
sin
x
x
=
1
]
and
B
=
lim
x
→
0
x
sin
(
1
x
)
⇒
B
=
0
Therefore,
A
=
1
and
B
=
0
is correct
The function
f
(
x
)
=
[
x
]
,
at
x
=
5
is:
Report Question
0%
left continuous
0%
right continuous
0%
continuous
0%
cannot be determined
Explanation
We know that the greatest integer function is discontinuous at integer values. For a value of
x
just smaller than 5, the greatest integer smaller than it is 4. So,
lim
x
→
5
−
[
x
]
=
4
f
(
x
=
5
)
=
[
5
]
=
5
For a value of
x
just bigger than 5, the greatest integer smaller than it is 5. So,
lim
x
→
5
+
[
x
]
=
5
Since
lim
x
→
5
−
f
(
x
)
≠
f
(
5
)
, the function is not left continuous.
Since
lim
x
→
5
+
f
(
x
)
=
f
(
5
)
, the function is right continuous.
Correct answer: Option B
f
(
x
)
=
{
2
x
−
1
i
f
x
>
2
k
i
f
x
=
2
x
2
−
1
i
f
x
<
2
is continuous at
x
=
2
then
k
=
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
Since, the function is continuous at x = 2,
lim
x
→
2
−
f
(
x
)
=
f
(
2
)
=
lim
x
→
2
+
f
(
x
)
lim
x
→
2
−
2
x
−
1
=
f
(
2
)
=
lim
x
→
2
+
x
2
−
1
k = 3
Hence, k = 3
Evaluate
lim
x
→
−
2
x
2
−
1
2
x
+
4
.
Report Question
0%
0
0%
1
0%
2
0%
∞
Explanation
Consider,
lim
x
→
−
2
+
x
2
−
1
2
x
+
4
Substitute
x
=
−
2
+
h
⟹
h
→
0
as
x
→
−
2
⟹
lim
x
→
−
2
+
x
2
−
1
2
x
+
4
=
lim
h
→
0
(
h
−
2
)
2
−
1
2
(
h
−
2
)
+
4
=
lim
h
→
0
h
2
−
4
h
+
3
2
h
=
∞
Hence,
lim
x
→
−
2
+
x
2
−
1
2
x
+
4
=
∞
If
f
(
x
)
=
sin
x
x
,
x
≠
0
is to be continuous at
x
=
0
then
f
(
0
)
=
Report Question
0%
0
0%
1
0%
−
1
0%
2
Explanation
From the property of limits,
L
i
m
x
→
0
s
i
n
x
x
=
1
Thus, for the function to be continuous at
x
=
0
f
(
0
)
=
1
Say true or false.
Every continuous function is always differentiable.
Report Question
0%
True
0%
False
Explanation
If a function
f
(
x
)
is differentiable at a point
a
, then it is continuous at the point
a
. But the converse is not true.
f
(
x
)
=
|
x
|
is contineous but not differentiable at
x
=
0.
Therefore, the given statement is false.
lim
x
→
∞
sin
x
equals
Report Question
0%
1
0%
0
0%
∞
0%
does not exist
Explanation
Let
L
=
lim
x
→
∞
sin
x
Assume
y
=
1
x
so as
x
→
∞
,
y
→
0
⇒
L
=
lim
y
→
0
sin
1
y
We know
sin
x
lie between -1 to 1
so let
p
=
sin
x
as
x
→
∞
Thus left hand limit
=
L
+
=
lim
y
→
0
+
sin
1
y
=
p
and right hand limit
=
L
−
=
lim
y
→
0
−
sin
1
y
=
−
p
Clearly L.H.L
≠
R.H.L
⇒
Required limit does not exist.
The function
f
(
x
)
=
|
x
−
3
|
x
−
3
at
x
=
3
, is
Report Question
0%
Left continuous
0%
Right continuous
0%
continuous
0%
discontinuous
Explanation
we just concentrate on the function
|
x
|
x
LHL = lim
x
→
0
−
|
x
|
x
=
−
x
x
=
−
1
RHL = lim
x
→
0
+
|
x
|
x
=
x
x
=
1
Let ,
y
=
x
−
3
Now, if
x
→
3
,
t
h
e
n
,
y
→
0
Applying the above derived result for y,
Since, the LHL is not equal to the RHL,
Hence the function is discontinuous at y = 0 or x = 3
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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