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Question
Mathematics
If (d y/d x)+2 y tan x= sin x, 0 < x < (π/2) and y((π/3))=0, then the maximum value of y(x) is
Q. If
d
x
d
y
+
2
y
tan
x
=
sin
x
,
0
<
x
<
2
π
and
y
(
3
π
)
=
0
, then the maximum value of
y
(
x
)
is
136
136
JEE Main
JEE Main 2022
Application of Derivatives
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A
8
1
B
4
3
C
4
1
D
8
3
Solution:
d
x
d
y
+
2
y
tan
x
=
sin
x
I.F
=
e
∫
2
t
a
n
x
d
x
=
e
l
n
(
s
e
c
x
)
2
=
sec
2
x
y
(
sec
2
x
)
=
∫
sin
x
sec
2
x
d
x
+
C
y
⋅
sec
2
x
=
sec
x
+
C
Put
x
=
3
π
,
y
=
0
y
=
cos
x
−
2
cos
2
x
=
8
1
−
2
(
cos
x
−
4
1
)
2
∴
y
m
a
x
=
8
1