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Test Index
JEE Main 29-Jan-2024 Shift 1 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 26
Total: 90
If the solution curve
y
=
y
(
x
)
of the differential equation
(
1
+
y
2
)
(
1
+
log
e
x
)
dx
+
x
d
y
=
0
,
x
>
0
passes through the point
(
1
,
1
)
and
y
(
e
)
=
α
−
tan
(
3
2
)
β
+
tan
(
3
2
)
, then
α
+
2
β
is
[29-Jan-2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
∫
(
1
x
+
ln
x
x
)
dx
+
∫
d
y
1
+
y
2
=
0
ln
x
+
(
ln
x
)
2
2
+
tan
−
1
y
=
C
Put
x
=
y
=
1
∴
C
=
π
4
⇒
ln
x
+
(
ln
x
)
2
2
+
tan
−
1
y
=
π
4
Put
x
=
e
⇒
y
=
tan
(
π
4
−
3
2
)
=
1
−
tan
3
2
1
+
tan
3
2
∴
α
=
1
,
β
=
1
⇒
α
+
2
β
=
3
© examsnet.com
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