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KEAM 2016 Math Paper
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© examsnet.com
Question : 97
Total: 120
If
s
=
sec
−
1
(
1
2
x
2
−
1
)
and
t
=
√
1
−
x
2
.
then
d
s
d
t
at
x
=
1
2
is
1
2
– 2
4
– 4
Validate
Solution:
Given,
s
=
sec
−
1
(
1
2
x
2
−
1
)
⇒
s
=
cos
−
1
(
2
x
2
−
1
)
[
∵
cos
−
1
x
=
sec
−
1
1
x
]
⇒
s
=
2
cos
−
1
x
[
∵
2
cos
−
1
x
=
cos
−
1
(
2
x
2
−
1
)
]
On differentiating w.r.t.
x
, we get
d
s
d
x
=
2
⋅
−
1
√
1
−
x
2
=
−
2
√
1
−
x
2
Also,
t
=
√
1
−
x
2
On differentiating w.r.t.
x
, we get
d
t
d
x
=
1
2
√
1
−
x
2
(
−
2
x
)
=
−
x
√
1
−
x
2
Now,
d
s
d
t
=
d
s
d
x
×
d
x
d
t
=
−
2
√
1
−
x
2
×
√
1
−
x
2
−
x
=
2
x
∴
(
d
s
d
t
)
x
=
1
2
=
2
1
/
2
=
4
© examsnet.com
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