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JEE Main 6 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 20
Total: 89
Let
f
:
(
−
∞
,
∞
)
−
{
0
}
⟶
R
be a differentiable function such that
f
′
(
1
)
=
lim
a
⟶
∞
a
2
f
(
1
a
)
. Then
lim
a
⟶
∞
a
(
a
+
1
)
2
tan
−
1
(
1
a
)
+
a
2
−
2
log
e
a
is equal to
[6 Apr 2024 Shift 1]
3
2
+
π
4
3
8
+
π
4
5
2
+
π
8
3
4
+
π
8
Validate
Solution:
👈: Video Solution
f
:
(
−
∞
,
∞
)
−
{
0
}
⟶
R
f
′
(
1
)
=
lim
a
⟶
∞
a
2
f
(
1
a
)
lim
a
⟶
∞
a
(
a
+
1
)
2
tan
−
1
(
1
a
)
+
a
2
−
2
ln
(
a
)
lim
a
⟶
∞
a
2
(
(
1
+
1
a
)
2
tan
−
1
(
1
a
)
+
1
−
2
a
2
ln
(
a
)
)
f
(
x
)
=
1
2
(
1
+
x
)
tan
−
1
(
x
)
+
1
−
2
x
2
ln
(
x
)
f
′
(
x
)
=
1
2
(
1
+
x
1
+
x
2
+
tan
−
1
(
x
)
+
4
x
ln
(
x
)
)
+
2
x
f
′
(
1
)
=
1
2
(
1
+
π
4
)
+
2
f
′
(
1
)
=
5
2
+
π
8
Ans. (3)
© examsnet.com
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