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JEE Main Math Class 11 Relations and Functions Part 1 Questions
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© examsnet.com
Question : 10
Total: 100
The domain of the function
f
(
x
)
=
cos
−
1
(
x
2
−
5
x
+
6
x
2
−
9
)
log
e
(
x
2
−
3
x
+
2
)
is :
[24-Jun-2022-Shift-1]
(
−
∞
,
1
)
∪
(
2
,
∞
)
(
2
,
∞
)
[
−
1
2
,
1
)
∪
(
2
,
∞
)
[
−
1
2
,
1
)
∪
(
2
,
∞
)
−
{
3
,
3
+
√
5
2
,
3
−
√
5
2
}
Validate
Solution:
👈: Video Solution
−
1
≤
x
2
−
5
x
+
6
x
2
−
9
≤
1
and
x
2
−
3
x
+
2
>
0
,
≠
1
(
x
−
3
)
(
2
x
+
1
)
x
2
−
9
≥
0
∣
5
(
x
−
3
)
x
2
−
9
≥
0
The solution to this inequality is
x
∈
[
−
1
2
,
∞
)
−
{
3
}
for
x
2
−
3
x
+
2
>
0
and
≠
1
x
∈
(
−
∞
,
1
)
∪
(
2
,
∞
)
−
{
3
−
√
5
2
,
3
+
√
5
2
}
Combining the two solution sets (taking intersection)
x
∈
[
−
1
2
,
1
)
∪
(
2
,
∞
)
−
{
3
−
√
5
2
,
3
+
√
5
2
}
© examsnet.com
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