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JEE Main Math Class 11 Trigonometric Functions Questions
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© examsnet.com
Question : 22
Total: 115
The set of all values of
λ
for which the equation
cos
2
2
x
−
2
s
i
n
4
x
−
2
cos
2
x
=
λ
[29-Jan-2023 Shift 2]
[
−
2
,
−
1
]
[
−
2
,
−
3
2
]
[
−
1
,
−
1
2
]
[
−
3
2
,
−
1
]
Validate
Solution:
λ
=
cos
2
2
x
−
2
s
i
n
4
x
−
2
cos
2
x
convert all in to
cos
x
.
λ
=
(
2
cos
2
x
−
1
)
2
−
2
(
1
−
cos
2
x
)
2
−
2
cos
2
x
=
4
cos
4
x
−
4
cos
2
x
+
1
−
2
(
1
−
2
cos
2
x
+
cos
4
x
)
−
2
cos
2
x
=
2
cos
4
x
−
2
cos
2
x
+
1
−
2
=
2
cos
4
x
−
2
cos
2
x
−
1
=
2
[
cos
4
x
−
cos
2
x
−
1
2
]
=
2
[
(
cos
2
x
−
1
2
)
2
−
3
4
]
λ
max
=
2
[
1
4
−
3
4
]
=
2
×
(
−
2
4
)
=
−
1
(max Value)
λ
min
=
2
[
0
−
3
4
]
=
−
3
2
(MinimumValue)
So, Range
=
[
−
3
2
,
−
1
]
© examsnet.com
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