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JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 97
Total: 124
Let
y
=
y
(
x
)
be the solution of the differential equation
x
tan
(
y
x
)
d
y
=
(
y
tan
(
y
x
)
−
x
)
d
x
−
1
≤
x
≤
1
,
y
(
1
2
)
=
π
6
.
Then the area of the region bounded by the curves
x
=
0
,
x
=
1
√
2
and
y
=
y
(
x
)
in the upper half plane is:
[20 Jul 2021 Shift 1]
1
8
(
π
−
1
)
1
12
(
π
−
3
)
1
4
(
π
−
2
)
1
6
(
π
−
1
)
Validate
Solution:
We have
d
y
d
x
=
x
(
y
x
.
tan
y
x
−
1
)
x
tan
y
x
∴
d
y
d
x
=
y
x
−
cot
(
y
x
)
Put
y
x
=
v
⇒
y
=
v
n
∴
d
y
d
x
=
v
+
n
d
v
d
x
Now, we get
v
+
n
d
v
d
x
=
v
−
cot
(
v
)
⇒
∫
(
tan
)
d
v
=
−
∫
d
x
x
∴
ln
|
sec
(
y
x
)
|
=
−
l
n
|
x
|
+
c
As
(
1
2
)
=
(
y
x
)
⇒
C
=
0
∴
sec
(
y
x
)
=
1
x
⇒
cos
(
y
x
)
=
x
∴
y
=
x
cos
−
1
(
x
)
So, required bounded area
=
1
∕
√
2
∫
0
x
(
cos
−
1
x
)
d
x
=
(
π
−
1
8
)
∴
option (1) is correct.
© examsnet.com
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