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JEE Main Math Class 11 Relations and Functions Part 1 Questions
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© examsnet.com
Question : 7
Total: 100
The domain of the function
cos
−
1
(
2
sin
−
1
(
1
4
x
2
−
1
)
π
)
is :
[29-Jun-2022-Shift-1]
R
−
{
−
1
2
,
1
2
}
(
−
∞
,
−
1
]
∪
[
1
,
∞
)
∪
{
0
}
(
−
∞
,
−
1
2
)
∪
(
1
2
,
∞
)
∪
{
0
}
(
−
∞
,
−
1
√
2
]
∪
[
1
√
2
,
∞
)
∪
{
0
}
Validate
Solution:
−
1
≤
2
s
i
n
−
1
(
1
4
x
2
−
1
)
π
≤
1
⇒
−
π
2
≤
s
i
n
−
1
(
1
4
x
2
−
1
)
≤
π
2
⇒
−
1
≤
1
4
x
2
−
1
≤
1
∴
1
4
x
2
−
1
+
1
≥
0
⇒
1
+
4
x
2
−
1
4
x
2
−
1
≥
0
⇒
4
x
2
4
x
2
−
1
≥
0
⇒
4
x
2
(
2
x
+
1
)
(
2
x
−
1
)
≥
0
.
.
.
.
.
.
(
1
)
∴
x
∈
(
−
α
,
−
1
2
)
∪
{
0
}
∪
(
1
2
,
α
)
And
1
4
x
2
−
1
−
1
≤
0
⇒
1
−
4
x
2
+
1
4
x
2
−
1
≤
0
⇒
2
−
4
x
2
4
x
2
−
1
≤
0
⇒
2
x
2
−
1
4
x
2
−
1
≥
0
⇒
(
√
2
x
+
1
)
(
√
2
x
−
1
)
(
2
x
+
1
)
(
2
x
−
1
)
≥
0
x
∈
(
−
α
,
−
1
√
2
)
∪
(
−
1
2
,
1
2
)
∪
(
1
√
2
,
α
)
From (3) and (4), we get
∴
x
∈
[
−
α
,
−
1
√
2
)
∪
[
1
√
2
,
α
)
∪
{
0
}
© examsnet.com
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