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JEE Main Math Class 11 Trigonometric Functions Questions
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© examsnet.com
Question : 1
Total: 115
Let
S
=
{
s
i
n
2
2
θ
:
(
s
i
n
4
θ
+
cos
4
θ
)
x
2
+
(
s
i
n
2
θ
)
x
+
.
(
s
i
n
6
θ
+
cos
6
θ
)
=
0
has real roots
}
. If
α
and
β
be the smallest and largest elements of the set
S
, respectively, then
3
(
(
α
−
2
)
2
+
(
β
−
1
)
2
)
equals.....
[4 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
D
=
(
s
i
n
2
θ
)
2
−
4
(
1
−
s
i
n
2
2
θ
2
)
(
1
−
3
4
s
i
n
2
2
θ
)
=
(
s
i
n
2
θ
)
2
−
4
(
1
−
5
4
s
i
n
2
2
θ
+
3
8
s
i
n
4
2
θ
)
D
=
−
3
2
s
i
n
4
2
θ
+
6
s
i
n
2
2
θ
−
4
>
0
3
s
i
n
4
2
θ
−
12
s
i
n
2
2
θ
+
8
<
0
s
i
n
2
2
θ
=
12
±
√
12
2
−
12.8
6
=
12
±
4
√
3
6
=
6
±
2
√
3
3
s
i
n
2
2
θ
=
2
±
2
√
3
,
but
s
i
n
2
2
θ
∈
[
0
,
1
]
∴
α
=
2
−
2
√
3
,
β
=
1
⟶
(
α
−
2
)
2
=
4
3
,
(
β
−
1
)
2
=
0
3
(
α
−
2
)
2
+
(
β
−
1
)
2
=
4
© examsnet.com
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