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Byju's Answer
Standard XII
Mathematics
Bijective Function
Which of the ...
Question
Which of the following functions are bijective?
A
f
:
R
→
[
0
,
∞
)
defined as
f
(
x
)
=
e
sgn
(
x
)
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B
f
:
[
3
,
4
]
→
[
4
,
6
]
defined by
f
(
x
)
=
|
x
−
1
|
+
|
x
−
2
|
+
|
x
+
3
|
+
|
x
−
4
|
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C
f
:
R
→
R
defined by
f
(
x
)
=
[
x
]
+
cos
(
π
[
x
]
)
where
[
.
]
denotes the greatest integer function
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D
f
:
R
→
R
defined as
f
(
x
)
=
min
{
x
+
2
,
−
2
x
+
4
}
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Solution
The correct option is
B
f
:
[
3
,
4
]
→
[
4
,
6
]
defined by
f
(
x
)
=
|
x
−
1
|
+
|
x
−
2
|
+
|
x
+
3
|
+
|
x
−
4
|
sgn
(
x
)
=
⎧
⎪
⎨
⎪
⎩
|
x
|
x
,
x
≠
0
0
,
x
=
0
sgn
(
x
)
=
⎧
⎨
⎩
−
1
,
x
<
0
0
,
x
=
0
1
,
x
>
0
∴
e
sgn
(
x
)
=
⎧
⎪ ⎪ ⎪
⎨
⎪ ⎪ ⎪
⎩
e
−
1
=
1
e
,
x
<
0
e
0
=
1
,
x
=
0
e
1
=
e
,
x
>
0
Neither one-one nor onto.
Hence,
e
sgn
(
x
)
is not bijective.
f
:
[
3
,
4
]
→
[
4
,
6
]
f
(
x
)
=
|
x
−
1
|
+
|
x
+
2
|
+
|
x
−
3
|
+
|
x
−
4
|
For
3
≤
x
≤
4
f
(
x
)
=
x
−
1
+
x
−
2
+
x
−
3
+
4
−
x
=
2
x
−
2
f
is one-one.
3
≤
x
≤
4
⇒
6
≤
2
x
≤
8
⇒
4
≤
2
x
−
2
≤
6
f
is onto.
It is bijective.
f
:
R
→
R
f
(
x
)
=
[
x
]
+
cos
π
[
x
]
If
0
≤
x
<
1
,
f
(
x
)
=
0
+
cos
π
(
0
)
=
1
Not one-one, so, not bijective.
f
:
R
→
R
f
(
x
)
=
min
{
x
+
2
,
−
2
x
+
4
}
Not one-one
Not onto.
So,
f
is not bijective.
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