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IIT JEE Advanced 2012 Paper 2
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© examsnet.com
Question : 58
Total: 60
Let f : (-1, 1)→R be such that f (cos 4θ) =
2
2
−
s
e
c
2
θ
for θ ∊
(
0
,
π
4
)
∪
(
π
4
,
π
2
)
. Then the value (s) of
f
(
1
3
)
is (are)
1
−
√
3
2
1
+
√
3
2
1
−
√
2
3
1
+
√
2
3
Validate
Solution:
For θ ∊
(
0
,
π
4
)
∪
(
π
4
,
π
2
)
Let cos 4θ =
1
3
.
⇒ cos 2θ = ±
√
1
+
c
o
s
4
θ
2
=
±
√
2
3
f
(
1
3
)
=
2
2
−
s
e
c
2
θ
=
2
c
o
s
2
θ
2
c
o
s
2
θ
−
1
=
1
+
1
c
o
s
2
θ
f
(
1
3
)
=
1
−
√
3
2
or
1
+
√
3
2
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