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JEE Main Math Class 12 Limits and Derivatives Part 2 Questions
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© examsnet.com
Question : 2
Total: 56
If
y
(
θ
)
=
2
cos
θ
+
cos
2
θ
cos
3
θ
+
4
cos
2
θ
+
5
cos
θ
+
2
, then at
θ
=
π
2
,
y
′
′
+
y
′
+
y
is equal to:
[5 Apr 2024 Shift 2]
3
2
1
1
2
2
Validate
Solution:
👈: Video Solution
y
=
2
cos
θ
+
2
cos
2
θ
−
1
4
cos
3
θ
−
3
cos
θ
+
8
cos
2
θ
−
4
+
5
cos
θ
+
2
y
=
(
2
cos
2
θ
+
2
cos
θ
−
1
)
(
2
cos
2
θ
+
2
cos
θ
−
1
)
(
2
cos
θ
+
2
)
y
=
1
2
(
1
1
+
cos
θ
)
⇒
θ
=
π
2
y
=
1
2
y
′
=
1
2
(
−
1
(
1
+
cos
θ
)
2
×
(
−
s
i
n
θ
)
)
⇒
θ
=
π
2
y
=
1
2
y
′
′
=
1
2
[
cos
θ
(
1
+
cos
θ
)
2
−
s
i
n
θ
(
2
)
(
1
+
cos
θ
)
(
−
s
i
n
θ
)
(
1
+
cos
θ
)
4
]
⇒
θ
=
π
2
y
=
1
© examsnet.com
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