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IIT JEE Advanced 2008 Paper 1
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Section:
Mathematics
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© examsnet.com
Question : 50
Total: 69
If 0 < x < 1, then
√
1
+
x
2
[
(
x
c
o
s
(
c
o
t
−
1
x
)
+
s
i
n
(
c
o
t
−
1
x
)
)
2
−
1
]
1
2
is equal to
[JEE Adv 2008 P1]
x
√
1
+
x
2
x
x
√
1
+
x
2
√
1
+
x
2
Validate
Solution:
We have
√
1
+
x
2
[
(
x
c
o
s
(
c
o
t
−
1
x
)
+
s
i
n
(
c
o
t
−
1
x
)
)
2
−
1
]
1
2
√
1
+
x
2
[
(
x
c
o
s
c
o
s
−
1
x
√
1
+
x
2
+
s
i
n
s
i
n
−
1
1
√
1
+
x
2
)
2
−
1
]
√
1
+
x
2
[
(
x
2
√
1
+
x
2
+
1
√
1
+
x
2
)
2
−
1
]
1
2
√
1
+
x
2
[
(
√
x
2
+
1
)
2
−
1
]
1
2
=
x
√
1
+
x
2
© examsnet.com
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