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Problem
Problem
If
f
(
x
)
f(x)
f
(
x
)
is defined on domain
[
0
,
1
]
[0, 1]
[
0
,
1
]
then
f
(
2
sin
x
)
f(2 \sin x)
f
(
2
sin
x
)
is defined on (a)
⋃
n
∈
I
{
[
2
n
π
,
2
n
π
+
π
6
]
∪
[
2
n
π
+
5
π
6
,
(
2
n
+
1
)
π
]
}
\bigcup_{n \in I} \left\{ \left[2n\pi, 2n\pi + \frac{\pi}{6}\right] \cup \left[2n\pi + \frac{5\pi}{6}, (2n+1)\pi\right] \right\}
⋃
n
∈
I
{
[
2
nπ
,
2
nπ
+
6
π
]
∪
[
2
nπ
+
6
5
π
,
(
2
n
+
1
)
π
]
}
(b)
⋃
n
∈
I
[
2
n
π
,
2
n
π
+
π
6
]
\bigcup_{n \in I} \left[2n\pi, 2n\pi + \frac{\pi}{6}\right]
⋃
n
∈
I
[
2
nπ
,
2
nπ
+
6
π
]
(c)
⋃
n
∈
I
[
2
n
π
+
5
π
6
,
(
2
n
+
1
)
π
]
\bigcup_{n \in I} \left[2n\pi + \frac{5\pi}{6}, (2n+1)\pi\right]
⋃
n
∈
I
[
2
nπ
+
6
5
π
,
(
2
n
+
1
)
π
]
(d) None of these
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