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Byju's Answer
Standard XII
Mathematics
Inverse of a Function
Consider fx...
Question
Consider
f
(
x
)
=
x
2
;
x
ϵ
Q
and
f
(
x
)
=
1
;
x
∉
Q
; and
g
(
x
)
=
[
x
]
;
where [ ] denotes greatest integer function. Then
A
f
(
x
)
is continuous at exactly two points
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B
g
(
x
)
discontinuous at all
x
ϵ
I
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C
f
o
g
(
x
)
is discontinuous at all
x
ϵ
Q
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D
g
o
f
(
x
)
is continuous everywhere except at
x
=
integer
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Solution
The correct options are
A
f
(
x
)
is continuous at exactly two points
B
g
(
x
)
discontinuous at all
x
ϵ
I
f
o
r
f
(
x
)
t
o
b
e
c
o
n
t
i
n
u
o
u
s
x
2
=
1
⇒
x
2
−
1
=
0
⇒
x
=
1
,
−
1
so f(x) is continuous at 2 points
g
(
x
)
=
[
x
]
g(x) is discontinuous at all integers because
l
i
m
x
→
k
−
g
(
x
)
=
k
−
1
l
i
m
x
→
k
+
g
(
x
)
=
k
Hence answer is A & B
Suggest Corrections
0
Similar questions
Q.
If [.] denotes greatest integer function and f(x) = [x]
{
s
i
n
π
[
x
+
1
]
+
s
i
n
π
[
x
+
1
]
1
+
[
x
]
}
,
then
Q.
Let
g
(
x
)
=
[
x
]
, where
[
.
]
represents greatest integer function. Then the function
f
(
x
)
=
(
g
(
x
)
)
2
−
g
(
x
)
is discontinuous at :
Q.
Let
f
(
x
)
=
[
x
]
and
g
(
x
)
=
s
g
n
(
x
)
(where
[
⋅
]
denotes greatest integer function), then discuss the continuity of
f
(
x
)
±
g
(
x
)
,
f
(
x
)
.
g
(
x
)
and
f
(
x
)
g
(
x
)
at
x
=
0
.
Q.
Given
f
(
x
)
=
|
x
−
1
|
+
|
x
+
1
|
. Then f(x) is
Q.
Assertion :If the series represented by function
f
(
x
)
=
x
2
+
x
4
+
x
6
+
x
8
+
.
.
.
converges, then function
g
(
x
)
=
[
x
]
(where [.] denotes the greatest integer function) is continuous at one fixed point of
f
(
x
)
. Reason:
f
(
x
)
=
x
⇒
x
2
+
x
−
1
=
0
which gives two fixed points.
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