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Byju's Answer
Standard XII
Mathematics
Definition of Functions
Domain of the...
Question
Domain of the function
f
(
x
)
=
1
x
+
2
(
sin
−
1
x
+
cos
−
1
x
)
+
1
√
4
−
x
2
is
Open in App
Solution
f
(
x
)
=
1
x
+
2
(
sin
−
1
x
+
cos
−
1
x
)
+
1
√
4
−
x
2
Since
sin
−
1
x
&
cos
−
1
x
is intrduced
x
ϵ
[
−
1
,
1
]
Also,
4
−
x
2
≥
0
but
x
≠
0
(
2
+
x
)
(
2
−
x
)
≥
0
(
x
+
2
)
(
x
−
2
)
≤
0
x
ϵ
[
−
2
,
2
]
∼
[
0
]
∴
domain
⇒
x
ϵ
[
−
1
,
0
)
∪
(
0
,
1
]
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