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UPSC NDA Math Model Paper 10
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© examsnet.com
Question : 96
Total: 120
Find the value of
d
d
x
(
sec
−
1
√
1
+
x
2
1
−
x
2
+
sin
−
1
√
1
−
x
2
1
+
x
2
)
π
∕
2
0
√
1
+
x
2
1
−
x
2
−
π
∕
2
Validate
Solution:
Given:
d
d
x
(
sec
−
1
√
1
+
x
2
1
−
x
2
+
sin
−
1
√
1
−
x
2
1
+
x
2
)
d
d
x
(
ocs
−
1
√
1
−
x
2
1
+
x
2
+
sin
−
1
√
1
−
x
2
1
+
x
2
)
[
∵
sec
−
1
x
=
cos
−
1
(
1
∕
x
)
]
We know that,
sin
−
1
x
+
cos
−
1
x
=
π
∕
2
=
d
d
x
(
π
2
)
=
0
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