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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 15
if the function
f
(
x
)
=
{
a
+
sin
−
1
(
x
+
b
)
,
x
≥
1
x
,
x
<
1
is differentiable at
x
=
1
, then
a
b
is equal to
Report Question
0%
1
0%
0
0%
−
1
0%
2
The function
f
(
x
)
=
x
1
+
|
x
|
is differentiable at which of the following?
Report Question
0%
Every where
0%
Everywhere except at
x
=
1
0%
Everywhere except at
x
=
0
0%
Everywhere except at
x
=
0
or
1
If
y
=
e
sin
−
1
(
t
2
−
1
)
&
x
=
e
sec
−
1
(
1
t
2
−
1
)
, then
d
y
d
x
is equal to
Report Question
0%
x
y
0%
−
y
x
0%
y
x
0%
−
x
y
If
f
(
x
)
=
s
i
n
−
1
[
2
x
1
+
x
2
]
,then
f
(
x
)
is differentiable on
Report Question
0%
[-1,1]
0%
R-{-1,1}
0%
R-(-1,1)
0%
None of these
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)
If
f
n
(
x
)
=
e
f
n
−
1
(
x
)
for all
n
ϵ
N
and
f
0
(
x
)
=
x
then
d
d
x
{
f
n
(
x
)
}
is equal to
Report Question
0%
f
n
(
x
)
.
d
d
x
{
f
n
−
1
(
x
)
}
0%
f
n
(
x
)
.
f
n
−
1
(
x
)
0%
f
n
(
x
)
.
f
n
−
1
(
x
)
.
.
.
.
.
f
2
(
x
)
.
f
1
(
x
)
0%
n
∏
i
=
1
f
i
(
x
)
Explanation
Hence, Option (A,C,D) is the correct answer.
The order of the differential equation of all circles whose radius is
4
, is?
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
Circle with radius
4
is represented as follows:-
(
x
−
h
)
2
+
(
y
−
k
)
2
=
4
2
.....
(
1
)
Differentiating w.r.t
x
2
(
x
−
h
)
+
2
(
y
−
k
)
d
y
d
x
=
0
⇒
(
x
−
h
)
+
(
y
−
k
)
d
y
d
x
=
0
.....
(
2
)
Again differentiating w.r.t
x
1
+
(
y
−
k
)
d
2
y
d
x
2
+
(
d
y
d
x
)
2
=
0
⇒
(
y
−
k
)
=
−
(
d
y
d
x
)
2
−
1
d
2
y
d
x
2
....
(
3
)
From
(
2
)
we get
⇒
(
x
−
h
)
=
−
(
y
−
k
)
d
y
d
x
⇒
(
x
−
h
)
=
−
{
(
d
y
d
x
)
2
−
1
d
2
y
d
x
2
}
d
y
d
x
∴
(
x
−
h
)
=
(
d
y
d
x
)
3
+
(
d
y
d
x
)
d
2
y
d
x
2
....
(
4
)
Substituting
(
4
)
and
(
3
)
in eq.
(
1
)
(
(
d
y
d
x
)
3
+
d
y
d
x
d
2
y
d
x
2
)
2
+
(
−
(
d
y
d
x
)
2
−
1
d
2
y
d
x
2
)
2
=
4
2
[
(
d
y
d
x
)
3
+
d
y
d
x
]
2
+
[
(
d
y
d
x
)
2
+
1
]
2
=
16
d
2
y
d
x
2
∴
Order of D.E of circle whose radius is
4
is
2
.
If
f
(
x
)
=
|
x
|
|
sin
x
|
, then
f
′
(
−
π
4
)
is equals
Report Question
0%
(
π
4
)
1
/
√
2
(
−
√
2
2
ln
4
π
−
2
√
2
π
)
0%
(
π
4
)
1
/
√
2
(
√
2
2
ln
4
π
+
2
√
2
π
)
0%
(
π
4
)
1
/
√
2
(
√
2
2
ln
π
4
+
2
√
2
π
)
0%
None of these.
Let
f
:
(
−
1
,
1
)
→
R
be a differentiable function satisfying
(
f
′
(
x
)
)
4
=
16
(
f
(
x
)
)
2
for all
x
∈
(
−
1
,
1
)
f
(
0
)
=
0
The number of such functions is
Report Question
0%
2
0%
3
0%
4
0%
more than
4
Explanation
Given:
(
f
′
(
x
)
)
4
=
16
(
f
(
x
)
)
2
for all
x
∈
(
–
1
,
1
)
f
(
0
)
=
0
⇒
(
f
′
(
x
)
)
2
=
±
4
f
(
x
)
⇒
f
′
(
x
)
=
±
2
√
±
f
(
x
)
(
i
)
Case
1
f
′
(
x
)
=
2
√
±
f
(
x
)
∫
d
(
f
(
x
)
)
f
(
x
)
=
∫
2
d
x
⇒
2
√
f
(
x
)
=
2
x
+
c
⇒
f
(
0
)
=
0
⇒
c
=
0
⇒
√
f
(
x
)
=
x
⇒
x
≥
0
f
(
x
)
=
x
2
,
0
≤
x
<
1
(
i
i
)
Case
2
f
′
(
x
)
=
−
2
√
f
(
x
)
⇒
√
f
(
x
)
=
−
x
⇒
x
≤
0
f
(
x
)
=
x
2
;
−
1
<
x
≤
0
(
i
i
i
)
Case
3
f
′
(
x
)
=
2
√
−
f
(
x
)
√
−
f
(
x
)
=
x
−
f
(
x
)
=
x
2
f
(
x
)
=
−
x
2
;
0
≤
x
<
1
(
i
v
)
Case
4
f
′
(
x
)
=
−
2
√
−
f
(
x
)
√
−
f
(
x
)
=
−
x
f
(
x
)
=
−
x
2
;
−
1
<
x
≤
0
(
v
)
Also, one singular solution of given differential equation is
f
(
x
)
=
0
,
−
1
<
x
<
1
Hence, there are more than
4
function possible
f
1
(
x
)
=
{
x
2
;
0
≤
x
<
1
−
x
2
;
−
1
<
x
<
0
f
2
(
x
)
=
{
−
x
2
;
0
≤
x
<
1
x
2
;
−
1
<
x
<
0
f
3
(
x
)
=
x
2
,
−
1
<
x
<
1
f
4
(
x
)
=
−
x
2
,
−
1
<
x
<
1
f
5
(
x
)
=
0
,
−
1
<
x
<
1
...
Hence there are more than
4
solutions
If
y
=
√
(
1
+
t
2
)
−
√
(
1
−
t
2
)
√
(
1
+
t
2
)
+
√
(
1
−
t
2
)
and
x
=
√
(
1
−
t
4
)
, then
d
y
d
x
Report Question
0%
−
1
t
2
{
1
+
√
1
−
t
4
}
0%
{
√
(
1
−
t
4
)
−
1
}
t
6
0%
1
t
2
{
1
+
√
(
1
−
t
4
)
}
0%
1
−
√
(
1
−
t
4
)
t
6
f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then
Report Question
0%
g(x) must be differentiable at x=a
0%
if g(x) is discontinuous , then f(a) =0
0%
f(a)
≠
0 then g(x) must be differentiable
0%
nothing can be said
Give that f(x) =xg(x) /
|
x
|
, g(0) = 0 and f(x) is continous at x=Then the value of f' (0)
Report Question
0%
Does not exist
0%
is -1
0%
is 1
0%
is 0
Let f(x)=
Report Question
0%
f' is differentiable
0%
f is differentiable
0%
f' is continuous
0%
f is continuous
If
f
(
x
)
=
{
x
log
cos
x
log
(
1
+
x
2
)
,
x
≠
0
0
,
x
=
0
then
Report Question
0%
f(x) is not continuous at x=0.
0%
f(x) is continuous at x=0.
0%
f(x) is continuous at x=0 but not differentiable at x=0.
0%
f(x) is differentiable at x=0.
Let f:
R
→
R
be a function such that f(x+y)= f(x)+f(y),
∀
x,y
ϵ
R
. If f(x) is differentiable at x=0, then
Report Question
0%
f(x) is differentiable only in a finite interval containing zero
0%
f(x) is continuous
∀
x
ϵ
R
0%
f(x) is constant
∀
x
ϵ
R
0%
f(x) is differentiable except at finitely many points
Let
f
(
x
)
be a function satisfying
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
and
f
(
x
)
=
x
g
(
x
)
∀
x
,
y
∈
R, where
g
(
x
)
is a continuous function then, which of the following is true?
Report Question
0%
f
′
(
x
)
=
g
′
(
0
)
0%
f
′
(
x
)
=
g
(
0
)
0%
f
(
x
)
=
g
(
0
)
0%
f
(
x
)
=
g
′
(
0
)
Which of the following is differentiable at x= 0
Report Question
0%
cos
(
|
x
|
)
+
|
x
|
0%
cos
(
|
x
|
)
−
|
x
|
0%
sin
(
|
x
|
)
+
|
x
|
0%
s
i
n
(
|
x
|
)
−
|
x
|
cos
|
x
|
is differentiable everywhere.
Report Question
0%
True
0%
False
If the function
f
:
[
0
,
8
]
→
R
is differentiable, then for
0
<
a
,
b
<
2
,
∫
8
0
f
(
t
)
d
t
is equal to
Report Question
0%
3
[
α
3
f
(
α
2
)
+
β
2
f
(
β
2
)
]
0%
3
[
α
3
f
(
α
)
+
β
3
f
(
β
)
]
0%
3
[
α
2
f
(
α
3
)
+
β
2
f
(
β
3
)
]
0%
3
[
α
2
f
(
α
2
)
+
β
2
f
(
β
2
)
]
Explanation
Let
g
(
x
)
=
∫
x
3
0
f
(
t
)
d
t
Now
∫
8
0
f
(
t
)
d
t
=
g
(
2
)
=
g
(
2
)
−
g
(
1
)
2
−
1
+
g
(
1
)
−
g
(
0
)
1
−
0
=
g
′
(
α
)
+
g
′
(
β
)
=
3
[
α
2
f
(
α
3
)
+
β
2
f
(
β
3
)
]
Let f(x) and g(x) be differentiable for
0
≤
x
≤
1
, such that f(0) such that f'(c)=2g'(c), then the value of g(1) must be
Report Question
0%
1
0%
3
0%
-2
0%
-1
f
(
x
)
=
x
2
+
x
g
′
(
1
)
+
g
′
′
(
2
)
and
g
(
x
)
=
f
(
1
)
x
2
+
x
f
′
(
x
)
+
f
′
′
(
x
)
The value of
g
(
0
)
is
Report Question
0%
0
0%
-3
0%
2
0%
None of these
Explanation
f
(
x
)
=
x
2
+
x
g
′
(
1
)
+
g
′
′
(
2
)
and
g
(
x
)
=
f
(
1
)
x
2
+
x
f
′
(
x
)
+
f
′
′
(
x
)
Here put
g
′
(
1
)
=
a
,
g
′
′
(
2
)
=
b
(
1
)
Then
f
(
x
)
=
x
2
+
a
x
+
b
,
f
(
1
)
=
1
+
a
+
b
⇒
f
′
(
x
)
=
2
x
+
a
f
′
′
(
x
)
=
2
∴
g
(
x
)
=
(
1
+
a
+
b
)
x
2
+
(
2
x
+
a
)
x
+
2
=
x
2
(
3
+
a
+
b
)
+
a
x
+
2
⇒
g
′
(
x
)
=
2
x
(
3
+
a
+
b
)
+
a
and
g
′
′
(
x
)
=
2
(
3
+
a
+
b
)
Hence,
g
′
(
1
)
=
2
(
3
+
a
+
b
)
+
a
(
2
)
g
′
′
(
2
)
=
2
(
3
+
a
+
b
)
(
3
)
From (1),(2) and (3) we have
a
=
2
(
3
+
a
+
b
)
+
a
and
b
=
2
(
3
+
a
+
b
)
⇒
3
+
a
+
b
=
0
and
b
+
2
a
+
6
=
0
Hence,
b
=
0
and
a
=
−
3
.
So,
f
(
x
)
=
x
2
−
3
x
and
g
(
x
)
=
−
3
x
+
2
g
(
x
)
=
−
3
x
+
2
⇒
g
(
0
)
=
2
f
(
x
)
is not invertible for
Report Question
0%
x
∈
[
−
π
2
−
tan
−
1
2
,
π
2
−
tan
−
1
2
]
0%
x
∈
[
tan
−
1
1
2
,
π
+
tan
−
1
1
2
]
0%
x
∈
[
π
+
cot
−
1
2
,
2
π
+
cot
−
1
2
]
0%
None of these
Explanation
f
(
x
)
=
sin
x
+
sin
x
∫
π
/
2
−
π
/
2
f
(
t
)
d
t
+
cos
x
∫
π
/
2
−
π
/
2
t
f
(
t
)
d
t
=
sin
x
(
1
+
∫
π
/
2
−
π
/
2
f
(
t
)
d
t
)
+
cos
x
∫
π
/
2
−
π
/
2
t
(
t
)
d
t
=
A
sin
x
+
B
cos
x
Thus,
A
=
1
+
∫
π
/
2
−
π
/
2
f
(
t
)
d
t
=
1
+
∫
π
/
2
−
π
/
2
(
A
sin
t
+
B
cos
t
)
d
t
=
1
+
2
B
∫
π
/
2
0
cos
t
d
t
=
∫
π
/
2
−
π
/
2
t
(
A
sin
t
+
B
cos
t
)
d
t
=
1
+
2
B
=
2
A
∫
π
/
2
0
t
f
(
t
)
d
t
=
2
A
[
−
t
cos
t
+
sin
t
]
π
/
2
0
⇒
B
=
2
A
From equations (1) and
(
2
)
,
we get
A
=
−
1
/
3
,
B
=
−
2
/
3
⇒
f
(
x
)
=
−
1
3
(
sin
x
+
2
cos
x
)
Thus, the range of
f
(
x
)
is
[
−
√
5
3
,
√
5
3
]
f
(
x
)
=
−
1
3
(
sin
x
+
2
cos
x
)
=
−
√
5
3
sin
(
x
+
tan
−
1
2
)
=
−
√
5
3
cos
(
x
−
tan
−
1
1
2
)
f
(
x
)
is invertible if
−
π
2
≤
x
+
tan
−
1
2
≤
π
2
⇒
−
π
2
−
tan
−
1
2
≤
x
≤
π
2
−
tan
−
1
2
or
0
≤
x
−
tan
−
1
1
2
≤
π
⇒
tan
−
1
1
2
≤
x
≤
π
+
tan
−
1
1
2
or
π
≤
x
−
tan
−
1
1
2
≤
2
π
⇒
x
∈
[
π
+
cot
−
1
2
,
2
π
+
cot
−
1
2
]
Let
f
(
0
,
∞
)
→
R
be a differentiable function such that
f
′
(
x
)
=
2
−
f
(
x
)
x
for all
x
ϵ
(
0
,
∞
)
and
f
(
1
)
≠
1
Then
Report Question
0%
l
i
m
x
→
0
+
f
′
(
1
x
)
=
1
0%
l
i
m
x
→
0
+
x
f
′
(
1
x
)
=
2
0%
l
i
m
x
→
0
+
x
2
f
′
(
1
x
)
=
0
0%
|
f
(
x
)
|
≤
2
for all
x
ϵ
(
0
,
2
)
0:0:1
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Incorrect : 0
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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