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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 15 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 15
L
t
x
→
π
4
√
2
−
cos
x
−
sin
x
(
4
x
−
π
)
2
=
?
Report Question
0%
1
16
√
2
0%
1
32
√
2
0%
1
16
0%
1
8
lim
x
→
0
1
−
cos
3
x
x
sin
2
x
=
Report Question
0%
1
/
2
0%
3
/
2
0%
3
/
4
0%
1
/
4
The value of
l
i
m
x
→
0
(
(
s
i
n
x
)
1
/
x
+
(
1
x
)
s
i
n
x
)
equals
Report Question
0%
0
0%
1
0%
∞
0%
-1
lim
x
→
0
ln
(
sin
3
x
)
ln
(
sin
x
)
is equal to
Report Question
0%
0
0%
1
0%
2
0%
none of these
l
i
m
x
→
0
1
e
2
t
a
n
(
π
4
+
x
)
1
/
x
Report Question
0%
0
0%
1
0%
-1
0%
e
The value of
θ
,
i
s
l
i
m
o
→
o
c
o
s
2
{
1
−
c
o
s
2
(
1
−
c
o
s
2
.
.
.
.
.
(
c
o
s
2
{
1
−
c
o
s
2
θ
}
)
)
}
s
i
n
(
π
(
√
θ
+
4
−
2
θ
)
Report Question
0%
√
2
4
0%
√
2
0%
1
0%
2
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
d
d
x
(
sin
5
x
sin
5
x
)
=
Report Question
0%
(
sin
4
x
sin
5
x
)
0%
5
(
sin
4
x
sin
5
x
)
0%
5
(
sin
4
x
sin
6
x
)
0%
−
5
(
sin
4
x
sin
6
x
)
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
h
→
0
sin
(
a
+
3
h
)
−
3
sin
(
a
+
2
h
)
+
3
sin
(
a
+
h
)
−
sin
a
h
3
is equal to
Report Question
0%
cos
a
0%
−
cos
a
0%
sin
a
0%
sin
a
cos
a
L
i
m
x
→
∞
(
sin
√
x
+
1
−
sin
√
x
)
=
Report Question
0%
2
0%
-2
0%
0
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
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
lim
x
→
π
/
2
[
x
tan
x
−
(
π
2
)
sec
x
]
is equal to
Report Question
0%
1
0%
-1
0%
0
0%
None of these
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
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
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
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
If x sin
(
α
+
y
)
=
s
i
n
y
a
n
d
y
′
=
m
(
x
2
+
)
n
x
+
,
then
Report Question
0%
m-n=1
0%
m+n=1
0%
m
2
+
n
2
=
1
0%
m=n
lim
x
→
∞
(
x
2
sin
(
1
x
)
−
x
1
−
|
x
|
)
=
Report Question
0%
0
0%
1
0%
-1`
0%
2
L
t
x
→
1
(
1
−
x
)
tan
π
x
=
Report Question
0%
−
1
0%
0
0%
1
0%
2
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
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
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
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
y
=
sec
(
tan
−
1
x
)
then at
x
=
1
value of
d
y
d
x
equals :
Report Question
0%
1
2
0%
1
0%
√
2
0%
1
√
2
0:0:4
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0
Answered
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Incorrect : 0
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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