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JEE Main 9 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 14
Total: 90
If
log
e
y
=
3
s
i
n
−
1
x
, then
(
1
−
x
)
2
y
′
′
−
x
y
′
at
x
=
1
2
is equal to :
[9 Apr 2024 Shift 2]
9
e
π
∕
6
3
e
π
∕
6
3
e
π
∕
2
9
e
π
∕
2
Validate
Solution:
👈: Video Solution
ln
(
y
)
=
3
s
i
n
−
1
x
1
y
⋅
y
′
=
3
(
1
√
1
−
x
2
)
⇒
y
′
=
3
y
√
1
−
x
2
at
x
=
1
2
⇒
y
′
=
3
e
3
(
π
6
)
√
3
2
=
2
√
3
e
π
2
⇒
y
′
′
=
3
(
√
1
−
x
2
y
′
−
y
1
2
√
1
−
x
2
(
−
2
x
)
(
1
−
x
2
)
)
⇒
(
1
−
x
2
)
y
′
′
=
3
(
3
y
+
x
y
√
1
−
x
2
)
↓
at
x
=
1
2
,
y
=
e
3
s
i
n
−
1
(
1
2
)
=
e
3
(
π
6
)
=
e
π
2
=
3
e
π
2
(
3
+
1
√
3
)
(
1
−
x
2
)
y
′
′
−
x
y
′
|
a
t
x
=
1
2
=
3
e
π
2
(
3
+
1
√
3
)
−
1
2
(
2
√
3
e
π
2
)
=
9
e
π
2
© examsnet.com
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