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Question
Mathematics
If f(x)= cot -1[ √(1+ sin x/1- sin x) ], 0 le x le (π /4), then f'( (π /4) ) is equal to
Q. If
f
(
x
)
=
cot
−
1
[
1
−
s
i
n
x
1
+
s
i
n
x
]
,
0
≤
x
≤
4
π
,
then
f
′
(
4
π
)
is equal to
1938
175
J & K CET
J & K CET 2010
Continuity and Differentiability
Report Error
A
2
1
50%
B
2
−
1
0%
C
1
+
x
2
1
−
x
2
0%
D
1
−
x
2
x
50%
Solution:
f
(
x
)
=
cot
−
1
[
1
−
s
i
n
x
1
+
s
i
n
x
]
⇒
f
′
(
x
)
=
1
+
(
1
−
s
i
n
x
1
+
s
i
n
x
)
−
1
.
2
1
−
s
i
n
x
1
+
s
i
n
x
1
(
1
−
s
i
n
x
)
2
[(
1
−
s
i
n
x
)
c
o
s
x
−
(
1
+
s
i
n
x
)
(
−
c
o
s
x
)
=
2.2
1
+
s
i
n
x
−
(
1
−
s
i
n
x
)
1
−
s
i
n
x
[
(
1
−
s
i
n
x
)
2
2
c
o
s
x
]
=
−
2
1
1
+
s
i
n
x
1
−
s
i
n
x
.
(
1
−
s
i
n
x
)
c
o
s
x
f
′
(
x
)
=
−
2
(
1
+
s
i
n
x
)
(
1
−
s
i
n
x
)
c
o
s
x
=
−
2.
c
o
s
x
c
o
s
x
=
−
2
1
⇒
f
′
(
4
π
)
=
−
2
1
Alternate
f
(
x
)
=
cot
−
1
[
1
−
s
in
x
1
+
s
i
n
x
]
f
(
x
)
=
cot
−
1
[
(
c
o
s
2
x
+
s
n
2
x
)
2
(
c
o
s
2
x
+
s
i
n
2
x
)
2
]
f
(
x
)
=
cot
−
1
(
c
o
s
2
x
−
s
i
n
2
x
c
o
s
2
x
+
s
i
n
2
x
)
f
(
x
)
=
cot
−
1
[
1
−
t
a
n
2
x
1
+
t
a
n
2
x
]
=
cot
−
1
[
tan
(
4
π
+
2
x
)
]
=
cot
−
1
[
cot
{
2
π
−
(
4
π
+
2
x
)
}
]
=
4
π
−
2
x
f
′
(
x
)
=
−
2
1
⇒
f
′
(
4
π
)
=
−
2
1