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JEE Advanced 2015 Paper 2
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Section:
Mathematics
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© examsnet.com
Question : 48
Total: 60
Let
f
:
ℝ
→
ℝ
be a continuous odd function,which vanishes exactly at one point and
f
(
1
)
=
1
2
.Suppose that
F
(
x
)
=
x
∫
−
1
f
(
t
)
d
t
for all
x
∊
[
−
1
,
2
]
and
G
(
x
)
=
x
∫
−
1
t
|
f
(
f
(
t
)
)
|
d
t
for all
x
∊
[
−
1
,
2
]
.If
l
i
m
x
→
1
F
(
x
)
G
(
x
)
=
1
14
then the value of
f
(
1
2
)
is
[JEE Adv 2015 P2]
Your Answer:
Validate
Solution:
G
(
1
)
=
x
∫
−
1
t
|
f
(
f
(
t
)
)
|
d
t
=
0
f
(
−
x
)
=
−
f
(
x
)
Given
f
(
1
)
=
1
2
l
i
m
x
→
1
F
(
x
)
G
(
x
)
=
l
i
m
x
→
1
F
(
x
)
−
F
(
1
)
x
−
1
G
(
x
)
−
G
(
1
)
x
−
1
=
f
(
1
)
|
f
(
f
(
1
)
)
|
=
1
14
⇒
1
2
|
f
(
1
2
)
|
=
1
14
⇒
f
(
1
2
)
=
7
© examsnet.com
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