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JEE Mains 29-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 7
Total: 90
Let the solution curve
y
=
y
(
x
)
of the differential equation
(
1
+
e
2
x
)
(
d
y
d
x
+
y
)
=
1
pass through the point
(
0
,
π
2
)
. Then,
lim
x
→
∞
e
x
y
(
x
)
is equal to :
[29-Jul-2022-Shift-1]
π
4
3
π
4
π
2
3
π
2
Validate
Solution:
D.E.
(
1
+
e
2
x
)
(
d
y
d
x
+
y
)
=
1
⇒
d
y
d
x
+
y
=
1
1
+
e
2
x
I.F.
=
e
∫
1
.
d
x
=
e
x
∴
Solution
e
x
y
(
x
)
=
∫
e
x
1
+
e
2
x
d
x
⇒
e
x
y
(
x
)
=
tan
−
1
(
e
x
)
+
C
∵
It passes through
(
0
,
π
2
)
,
C
=
π
2
−
π
4
=
π
4
∴
lim
x
→
∞
e
x
y
(
x
)
=
lim
x
→
∞
tan
−
1
(
e
x
)
+
π
4
=
3
π
4
© examsnet.com
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