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JEE Advanced 2015 Paper 2 Mathematics Q7
Mathematics
Problem
Let
f
:
R
→
R
f:\mathbb R\to\mathbb R
f
:
R
→
R
be a continuous odd function, which vanishes exactly at one point and
f
(
1
)
=
1
2
f(1)=\dfrac12
f
(
1
)
=
2
1
. Suppose that
F
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
F(x)=\displaystyle\int_{-1}^{x}f(t)\,dt
F
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
for all
x
∈
[
−
1
,
2
]
x\in[-1,2]
x
∈
[
−
1
,
2
]
and
G
(
x
)
=
∫
−
1
x
t
∣
f
(
f
(
t
)
)
∣
d
t
G(x)=\displaystyle\int_{-1}^{x}t|f(f(t))|\,dt
G
(
x
)
=
∫
−
1
x
t
∣
f
(
f
(
t
))
∣
d
t
for all
x
∈
[
−
1
,
2
]
x\in[-1,2]
x
∈
[
−
1
,
2
]
. If
lim
x
→
1
F
(
x
)
G
(
x
)
=
1
14
\displaystyle\lim_{x\to1}\dfrac{F(x)}{G(x)}=\dfrac1{14}
x
→
1
lim
G
(
x
)
F
(
x
)
=
14
1
, then the value of
f
(
1
2
)
f\left(\dfrac12\right)
f
(
2
1
)
is
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