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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 9
If
f
(
x
)
is a differentiable function
∀
x
ϵ
R
so that,
f
(
2
)
=
4
,
f
′
(
x
)
≥
5
∀
x
ϵ
[
2
,
6
]
, then,
f
(
6
)
is :
Report Question
0%
≥
24
0%
≤
24
0%
≥
9
0%
≤
9
If
x
=
sin
−
1
(
2
θ
1
+
θ
2
)
,
y
=
sec
−
1
√
1
+
θ
2
, then
d
y
d
x
=
Report Question
0%
−
1
2
0%
1
2
0%
−
2
0%
2
If
f
(
x
)
=
0
for
x
<
0
and
f
(
x
)
is differential at
x
=
0
then for
x
≥
0
,
f
(
x
)
may be
Report Question
0%
x
2
0%
x
0%
−
x
0%
−
x
3
/
2
If
f
(
x
)
=
{
m
x
−
1
,
x
≤
5
3
x
−
5
,
x
>
5
is continuous then value of m is:
Report Question
0%
11
5
0%
5
11
0%
5
3
0%
3
5
Explanation
f
(
x
)
{
m
x
−
1
if
x
≤
5
3
x
−
5
if
x
>
5
Since the function is continuous at
x
=
5
.
Therefore,
lim
x
→
5
−
f
(
x
)
=
lim
x
→
5
+
f
(
x
)
=
f
(
5
)
lim
x
→
5
−
(
m
x
−
1
)
=
lim
x
→
5
+
(
3
x
−
5
)
5
m
−
1
=
15
−
5
5
m
=
10
+
1
⇒
m
=
11
5
Hence the value of
m
is
11
5
.
If
y
=
t
a
n
−
1
(
a
x
−
b
b
x
+
a
)
, the value of
d
y
d
x
is
Report Question
0%
a
1
+
x
2
0%
1
1
+
x
2
0%
−
b
1
+
x
2
0%
b
1
+
x
2
Let
f
:
[
0
,
2
]
→
R
be a twice differentiable function such that
f
"
(
x
)
>
0
, for all
x
∈
(
0
,
2
)
If
ϕ
(
x
)
=
f
(
x
)
+
f
(
2
−
x
)
, then
ϕ
is:
Report Question
0%
decreasing on
(
0
,
2
)
0%
decreasing on
(
0
,
1
)
and increasing on
(
1
,
2
)
0%
increasing on
(
0
,
2
)
0%
increasing on
(
0
,
1
)
and decreasing on
(
1
,
2
)
Explanation
ϕ
(
x
)
=
f
(
x
)
+
f
(
2
−
x
)
ϕ
′
(
x
)
=
f
′
(
x
)
−
f
′
(
2
−
x
)
...(1)
Since
f
"
(
x
)
>
0
⇒
f
′
(
x
)
i
s
i
n
c
r
e
a
s
i
n
g
∀
x
∈
(
0
,
2
)
case I when
x
>
2
−
x
⇒
x
>
1
⇒
ϕ
′
(
x
)
>
0
∀
x
∈
(
1
,
2
)
∴
ϕ
(
x
)
i
s
i
n
c
r
e
a
s
i
n
g
o
n
(
1
,
2
)
case II: when
x
<
2
−
x
⇒
x
<
1
⇒
ϕ
′
(
x
)
<
0
∀
x
∈
(
0
,
1
)
∴
ϕ
(
x
)
is decreasing on
(
0
,
1
)
If
f
(
x
)
=
|
cos
x
−
sin
x
|
, then
f
′
(
π
6
)
equal to?
Report Question
0%
−
1
2
(
1
+
√
3
)
0%
1
2
(
1
+
√
3
)
0%
−
1
2
(
1
−
√
3
)
0%
1
2
(
1
−
√
3
)
If
f
(
x
)
=
{
l
o
g
e
x
x
−
1
x
≠
1
k
x
=
1
continuous at
x
=
1
, then the value of k is?
Report Question
0%
e
0%
1
0%
−
1
0%
0
Explanation
Given
f
(
x
)
=
{
l
o
g
e
x
x
−
1
x
≠
1
k
x
=
1
we know that at
x
=
1
is continuous so
k
=
lim
x
→
1
l
o
g
e
x
x
−
1
It is
0
0
form so by applying L-hospital rule
=
lim
x
→
1
1
/
x
1
=
1
.
Let
f
(
x
)
=
15
−
|
x
−
10
|
;
x
∈
R
. Then the set of all values of x, at which the function,
g
(
x
)
=
f
(
f
(
x
)
)
is not differentiable, is?
Report Question
0%
{
5
,
10
,
15
,
20
}
0%
{
10
,
15
}
0%
{
5
,
10
,
15
}
0%
{
10
}
Explanation
f
(
x
)
=
15
−
|
x
−
10
|
,
x
∈
R
f
(
f
(
x
)
)
=
15
−
|
f
(
x
)
−
10
|
=
15
−
|
15
−
|
x
−
10
|
−
10
|
=
15
−
|
5
−
|
x
−
10
|
|
x
=
5
,
10
,
15
are points of non differentiability
Aliter:
At
x
=
10
f
(
x
)
is no differentiable
also, when
15
−
|
x
−
10
|
=
10
⇒
x
=
5
,
15
∴
non differentiability points are
{
5
,
10
,
15
}
.
If
f
(
x
)
=
|
√
1
+
k
x
−
√
1
−
k
x
i
f
−
1
≤
x
<
0
2
x
+
1
x
−
1
i
f
0
≤
x
≤
1
|
is continuous at
x
=
0
, then the value of k is?
Report Question
0%
k
=
1
0%
k
=
−
1
0%
k
=
0
0%
k
=
2
Explanation
lim
x
→
0
−
√
1
+
K
x
−
√
1
−
K
x
x
=
lim
x
→
0
+
m
K
x
x
(
√
1
−
K
x
+
√
x
)
lim
x
→
0
+
2
x
+
1
x
−
1
=
−
1
=
K
(after rationlation)
∴
K
=
−
1
.
If
f
(
x
)
=
{
sin
(
p
+
1
)
x
+
sin
x
x
,
x
<
0
q
,
x
=
0
√
x
2
+
x
−
√
x
x
3
/
2
,
x
>
0
is continuous at
x
=
0
the
(
p
,
q
)
is
Report Question
0%
(
−
1
2
,
−
3
2
)
0%
(
3
2
,
1
2
)
0%
(
1
2
,
3
2
)
0%
(
−
3
2
,
1
2
)
If
x
=
3
sin
t
,
y
=
3
cos
t
,
find
d
y
d
x
at
t
=
π
3
Report Question
0%
3
0%
0
0%
−
√
3
0%
1
Explanation
x
=
3
sin
t
d
x
d
t
=
3
cos
t
y
=
3
cos
t
d
y
d
t
=
−
3
sin
t
Thus,
d
y
d
x
=
−
tan
t
At
t
=
π
3
,
d
y
d
x
=
−
√
3
Let
f
(
x
)
=
(
x
+
|
x
|
)
|
x
|
. Then, for all x.
Report Question
0%
f
is continuous
0%
f
is differentiable for some x
0%
f
′
is continuous
0%
f
″
is continuous
Explanation
Consider the given function.
f
(
x
)
=
(
x
+
|
x
|
)
|
x
|
f
(
x
)
=
{
(
x
+
x
)
x
,
x
>
0
(
x
−
x
)
x
,
x
<
0
f
(
x
)
=
{
2
x
2
,
x
>
0
0
,
x
<
0
f
′
(
x
)
=
{
4
x
,
x
>
0
0
,
x
<
0
Now,
lim
x
→
0
+
2
x
2
=
lim
x
→
0
−
0
=
0
Since, it is polynomial function.
Hence, it is continuous and differentiable at
0
.
The set of points where the function
f
(
x
)
=
x
|
x
|
is differentiable is?
Report Question
0%
(
−
∞
,
∞
)
0%
(
−
∞
,
0
)
∪
(
0
,
∞
)
0%
(
0
,
∞
)
0%
[
0
,
∞
]
Explanation
Consider the given function.
f
(
x
)
=
x
|
x
|
f
(
x
)
=
{
x
2
,
x
>
0
−
x
2
,
x
<
0
0
x
=
0
Since, it is polynomial function.
Hence, it is differentiable at
(
−
∞
,
∞
)
.
Differentiate the following function with respect to x.
If for
f
(
x
)
=
λ
x
2
+
μ
x
+
12
,
f
′
(
14
)
=
15
and
f
′
(
2
)
=
11
, then find
λ
and
μ
.
Report Question
0%
λ
=
1
,
μ
=
6
.
0%
λ
=
1
,
μ
=
7
.
0%
λ
=
6
,
μ
=
7
.
0%
λ
=
6
,
μ
=
1
.
Explanation
Given function
f
(
x
)
=
λ
x
2
+
μ
x
+
12
⟹
f
′
(
x
)
=
d
d
x
(
λ
x
2
+
μ
x
+
12
)
f
′
(
x
)
=
2
λ
x
+
μ
f
′
(
2
)
=
2
λ
(
2
)
+
μ
⟹
4
λ
+
μ
=
11
⋯
(
1
)
f
′
(
4
)
=
2
λ
(
4
)
+
μ
⟹
8
λ
+
μ
=
15
⋯
(
2
)
(
2
)
−
(
1
)
8
λ
−
4
λ
+
μ
−
μ
=
15
−
11
4
λ
=
4
⟹
λ
=
1
Put it in
(
1
)
⟹
4
(
4
)
+
μ
=
11
μ
=
11
−
4
μ
=
7
If
f
(
9
)
=
9
,
f
′
(
9
)
=
0
, then
lim
x
→
9
√
f
(
x
)
−
3
√
x
−
3
is equal to
Report Question
0%
0
0%
f
(
0
)
0%
f
′
(
3
)
0%
f
(
9
)
0%
1
Explanation
lim
x
→
9
√
f
(
x
)
−
3
√
x
−
3
[
0
0
f
r
o
m
]
=
lim
x
→
9
f
′
(
x
)
2
√
f
(
x
)
1
2
√
x
=
√
x
f
′
(
x
)
√
f
(
x
)
=
√
9
f
′
(
9
)
√
f
(
9
)
=
3
×
0
3
=
0
Let
f
:
R
→
R
be a continuous function such that
f
(
x
2
)
=
f
(
x
3
)
for all
x
∈
R
. Consider the following statements.
I. f is an odd function.
II. f is an even function.
III. f is differentiable everywhere
Report Question
0%
I is true and III is false
0%
II is true and III is false
0%
both I and III are true
0%
both II and III are true
Explanation
f
:
R
→
R
be a continuous function such that
f
(
x
2
)
=
f
(
x
3
)
..........
(
1
)
for all
x
∈
R
Put
x
=
−
x
f
(
x
2
)
=
f
(
−
x
3
)
From
(
1
)
we have
f
(
x
3
)
=
f
(
−
x
3
)
Put
x
3
=
t
we have
f
(
t
)
=
f
(
−
t
)
⇒
f
(
x
)
is an even function.
(
i
i
)
Now take
x
3
=
t
⇒
f
(
t
2
3
)
=
f
(
t
)
Put
t
=
t
2
3
⇒
f
(
(
t
2
3
)
2
)
=
f
(
t
)
⇒
f
(
t
)
=
f
(
t
2
3
)
=
f
(
(
t
2
3
)
2
)
=
f
(
(
t
2
3
)
3
)
=
.
.
.
.
=
f
(
(
t
2
3
)
n
)
This is true for all
t
∈
R
and any
n
∈
I
Hence if we take
n
→
∞
,
(
2
3
)
n
→
0
Then
f
(
t
)
=
f
(
t
∘
)
=
1
⇒
f
(
x
)
is a constant function, hence it is differentiable everywhere.
Let
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
for all
x
and
y
. If
f
(
0
)
=
1
,
f
(
3
)
=
3
and
f
′
(
0
)
=
11
, then
f
′
(
3
)
is equal to
Report Question
0%
11
0%
22
0%
33
0%
44
Explanation
We have,
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
⇒
f
′
(
3
)
=
lim
h
→
0
f
(
3
+
h
)
−
f
(
3
)
h
=
lim
h
→
0
f
(
3
)
f
(
h
)
−
f
(
3
+
0
)
h
=
lim
h
→
0
f
(
3
)
f
(
h
)
−
f
(
3
)
f
(
0
)
h
=
f
(
3
)
lim
h
→
0
f
(
h
)
−
f
(
0
)
h
=
f
(
3
)
lim
h
→
0
f
(
0
+
h
)
−
f
(
0
)
h
=
f
(
3
)
f
′
(
0
)
=
3
×
11
=
33
Let
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
and
f
(
x
)
=
1
+
sin
(
3
x
)
g
(
x
)
, where
g
is differentiable. The
f
′
(
x
)
is equal to
Report Question
0%
3
f
(
x
)
0%
g
(
0
)
0%
3
f
(
x
)
g
(
0
)
0%
3
g
(
x
)
Explanation
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
f
(
x
)
f
(
h
)
−
f
(
x
)
h
=
f
(
x
)
lim
h
→
0
(
1
+
sin
3
h
(
g
(
h
)
)
−
1
h
)
3
f
(
x
)
lim
h
→
0
sin
3
h
3
h
lim
h
→
0
g
(
h
)
=
3
f
(
x
)
×
1
×
g
(
0
)
=
3
f
(
x
)
g
(
0
)
If
y
=
tan
−
1
(
a
cos
x
−
b
sin
x
b
cos
x
+
a
sin
x
)
then
d
y
d
x
=
?
Report Question
0%
a
b
0%
−
b
a
0%
1
0%
−
1
Explanation
Let,
a
b
=
tan
θ
y
=
tan
−
1
(
a
cos
x
−
b
sin
x
b
cos
x
+
a
sin
x
)
=
tan
−
1
(
a
b
−
tan
x
1
+
a
b
tan
x
)
=
tan
−
1
(
tan
θ
−
tan
x
1
+
tan
θ
tan
x
)
=
tan
−
1
tan
(
θ
−
x
)
=
θ
−
x
=
(
tan
−
1
a
b
−
x
)
.
∴
d
y
d
x
=
−
1
.
If
y
=
tan
−
1
{
cos
x
+
sin
x
cos
x
−
sin
x
}
then
d
y
d
x
=
?
Report Question
0%
1
0%
−
1
0%
1
2
0%
−
1
2
If
y
=
sin
−
1
(
3
x
−
4
x
3
)
then
d
y
d
x
=
?
Report Question
0%
3
√
1
−
x
2
0%
−
4
√
1
−
x
2
0%
3
√
1
+
x
2
0%
n
o
n
e
o
f
t
h
e
s
e
If
f
(
x
)
=
s
i
n
1
(
2
x
1
+
x
2
)
, then
Report Question
0%
f
is derivative for all
x
with ,
|
x
|
<
1
0%
f
is not derivative at
x
=
1
0%
f
is not derivable at
x
=
−
1
0%
f
is derivative for all
x
with ,
|
x
|
>
1
If
f
(
x
)
=
{
x
2
(
s
g
n
[
x
]
)
+
{
x
}
)
,
0
≤
x
<
2
sin
x
+
|
x
−
3
|
,
2
≤
x
<
4
were [ ] and { } represent the greatest integer and the fractional part function, respectively.
Report Question
0%
f (x) is differentiable at x=1.
0%
f(x) is continuous but non-differentiable at x=1.
0%
f(x) is non-differentiable at x=2.
0%
f(x) is non-differentiable at x=2.
If
y
=
cos
−
1
x
3
then
d
y
d
x
=
?
Report Question
0%
−
1
√
1
−
x
6
0%
−
3
x
2
√
1
−
x
6
0%
−
3
x
2
√
1
−
x
6
0%
n
o
n
e
o
f
t
h
e
s
e
If
y
=
cos
−
1
(
4
x
3
−
3
x
)
then
d
y
d
x
=
?
Report Question
0%
3
√
1
−
x
2
0%
−
3
√
1
−
x
2
0%
4
√
1
−
x
2
0%
−
4
(
3
x
2
−
1
)
Which of the following statement is always true ([ .] represents the greatest integer function)
Report Question
0%
If f(x) is discontinuous, then
|
f
(
x
)
|
is discontinuous
0%
If f(x) is discontinuous , then f
(
|
x
|
)
is discontinuous
0%
f(x) =
[
g
(
x
)
]
is discontinuous when g(x) is an integer
0%
None of these
The largest value of c such that there exists a differential function
h
(
x
)
f
o
r
−
c
<
x
<
c
that is a solution of
y
1
=
1
+
y
2
with h(0) = 0 is
Report Question
0%
2
π
0%
π
0%
π
2
0%
π
4
Which of the following function is thrice differentiable at x=0?
Report Question
0%
f
(
x
)
=
|
x
3
|
0%
f
(
x
)
=
x
3
|
x
|
0%
f
(
x
)
=
|
x
|
sin
3
x
0%
f
(
x
)
=
x
|
tan
3
x
|
If
y
=
tan
−
1
(
sec
x
+
6
tan
x
)
then
d
y
d
x
=
?
Report Question
0%
1
2
0%
−
1
2
0%
1
0%
n
o
n
e
o
f
t
h
e
s
e
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Answered
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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