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Byju's Answer
Standard XIII
Mathematics
Differentiability
Let gx=2x+1, ...
Question
Let
g
(
x
)
=
⎧
⎪
⎨
⎪
⎩
2
(
x
+
1
)
,
−
∞
<
x
≤
−
1
√
1
−
x
2
,
−
1
<
x
<
1
∣
∣
∣
∣
|
x
|
−
1
∣
∣
−
1
∣
∣
,
1
≤
x
<
∞
.
Then
A
g
(
x
)
is non-differentiable at exactly three points
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B
g
(
x
)
is continuous in
(
−
∞
,
1
]
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C
g
(
x
)
is differentiable in
(
−
∞
,
−
1
)
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D
g
(
x
)
has finite type of discontinuity at
x
=
1
, but continuous at
x
=
−
1
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Solution
The correct option is
D
g
(
x
)
has finite type of discontinuity at
x
=
1
, but continuous at
x
=
−
1
From the above graph, we can verify the alternatives.
Suggest Corrections
0
Similar questions
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
Domain of
f
(
g
(
x
)
)
is
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
Range of the function
f
(
g
(
x
)
)
is
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
The number of roots of the equation
f
(
g
(
x
)
)
=
2
is
Q.
Let
E
1
=
{
x
∈
R
:
x
≠
1
and
x
x
−
1
>
0
}
and
E
2
=
{
x
∈
E
1
:
sin
−
1
(
log
e
(
x
x
−
1
)
)
is a real number
}
.
(
Here, the inverse trigonometric function
sin
−
1
x
assumes values in
[
−
π
2
,
π
2
]
.
)
Let
f
:
E
1
→
R
be the function defined by
f
(
x
)
=
log
e
(
x
x
−
1
)
and
g
:
E
2
→
R
be the function defined by
g
(
x
)
=
sin
−
1
(
log
e
(
x
x
−
1
)
)
.
List I
List II
P
.
The range of
f
is
1.
(
−
∞
,
1
1
−
e
]
∪
[
e
e
−
1
,
∞
)
Q
.
The range of
g
contains
2.
(
0
,
1
)
R
.
The domain of
f
contains
3.
[
−
1
2
,
1
2
]
S
.
The domain of
g
is
4.
(
−
∞
,
0
)
∪
(
0
,
∞
)
5.
(
−
∞
,
e
e
−
1
]
6.
(
−
∞
,
0
)
∪
(
1
2
,
e
e
−
1
]
The correct option is:
Q.
Let
g
(
x
)
=
⎧
⎪
⎨
⎪
⎩
2
(
x
+
1
)
,
−
∞
<
x
≤
−
1
√
1
−
x
2
,
−
1
<
x
<
1
|
x
+
1
|
,
1
≤
x
<
∞
then
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