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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 16 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 16
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
If
x
=
3
cos
θ
−
2
cos
3
θ
and
y
=
3
sin
θ
−
2
sin
3
θ
, then
d
y
d
x
=
Report Question
0%
sin
θ
0%
cos
θ
0%
tan
θ
0%
cot
θ
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
If
y
=
e
sin
2
x
+
sin
4
x
+
sin
6
x
+
.
.
.
.
+
∞
, then
d
y
d
x
=
?
Report Question
0%
2
tan
x
sec
2
x
e
tan
2
x
0%
2
sec
2
x
e
tan
2
x
0%
sec
2
x
e
tan
2
x
0%
tan
x
sec
x
e
tan
2
x
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
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
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
d
d
x
(
3
cos
(
π
6
+
x
0
)
−
4
cos
3
(
π
6
+
x
0
)
)
=
.
.
.
.
Report Question
0%
cos
(
3
x
0
)
0%
π
60
sin
(
3
x
0
)
0%
π
60
cos
(
3
x
0
)
0%
−
π
60
sin
(
3
x
0
)
Explanation
we know that
c
o
s
(
3
x
)
=
4
c
o
s
3
x
−
3
c
o
s
x
so
d
d
x
(
−
cos
3
(
π
6
+
x
0
)
)
=
d
d
x
(
−
cos
(
π
2
+
3
x
0
)
)
=
d
d
x
(
sin
3
x
π
180
)
=
3
π
180
cos
(
3
x
0
)
=
π
60
cos
(
3
x
0
)
Value of
L
=
lim
n
→
∞
n
1
4
[
1.
(
n
∑
k
=
1
k
)
+
2.
(
n
−
1
∑
k
=
1
k
)
+
3.
(
n
−
2
∑
k
=
1
k
)
+
.
.
.
+
n
.1
]
is
Report Question
0%
1
/
24
0%
1
/
12
0%
1
/
6
0%
1
/
3
Explanation
If
y
=
4
l
o
g
2
s
i
n
x
+
9
l
o
g
3
c
o
s
x
then
d
y
d
x
=
Report Question
0%
0
0%
1
0%
-1
0%
2
Let
x
c
o
s
y
+
y
c
o
x
x
=
5
, Then
Report Question
0%
a
t
x
=
0
,
y
=
0
,
y
′
=
0
0%
a
t
x
=
0
,
y
=
1
,
y
′
=
0
0%
a
t
x
=
y
,
y
=
1
,
y
′
=
−
1
0%
a
t
x
=
1
,
y
=
0
,
y
′
=
1
The value of
lim
n
∞
1
n
2
{
s
i
n
3
π
4
n
+
2
s
i
n
3
2
π
4
n
+
.
.
.
+
n
s
i
n
3
n
π
4
n
}
is equal to
Report Question
0%
√
2
9
π
2
(
52
−
15
π
)
0%
2
9
π
2
(
52
−
15
n
)
0%
1
9
π
2
(
15
n
−
15
)
0%
None of these
If
lim
x
→
0
x
n
−
sin
x
n
x
−
sin
n
x
is non-zero finite, then
n
must be equal
Report Question
0%
4
0%
1
0%
2
0%
3
If
L
=
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
3
=
∞
Equation
a
x
2
+
b
x
+
c
=
0
has
Report Question
0%
real and equal roots
0%
complex roots
0%
unequal positive real roots
0%
unequal roots
lim
x
→
∞
2
+
2
x
+
sin
2
x
(
2
x
+
sin
2
x
)
e
sin
x
is equal to
Report Question
0%
0
0%
1
0%
-1
0%
Does not exists
lim
x
→
π
/
2
sin
(
x
cos
x
)
cos
(
x
sin
x
)
is equal to
Report Question
0%
0
0%
p/2
0%
p
0%
2p
The value of
lim
x
→
0
√
1
2
(
1
−
cos
2
x
)
x
is
Report Question
0%
1
0%
-1
0%
0
0%
None of these
Solution of the equation
d
y
d
x
+
1
x
tan
y
=
1
x
2
tan
y
sin
y
is
Report Question
0%
2
x
=
s
i
n
y
(
x
)
0%
2
x
=
s
i
n
y
(
1
+
c
x
2
)
0%
2
x
+
s
i
n
y
(
1
+
c
x
2
)
0%
None of these
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0
Answered
1
Not Answered
20
Not Visited
Correct : 0
Incorrect : 0
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