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Problem
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) = \frac{1}{2}
f
(
1
)
=
2
1
. Suppose that
F
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
for all
x
∈
[
−
1
,
2
]
F(x) = \int_{-1}^{x} f(t)\,dt \text{ for all } x \in [-1, 2]
F
(
x
)
=
∫
−
1
x
f
(
t
)
d
t
for all
x
∈
[
−
1
,
2
]
and
G
(
x
)
=
∫
−
1
x
t
∣
f
(
f
(
t
)
)
∣
d
t
for all
x
∈
[
−
1
,
2
]
.
G(x) = \int_{-1}^{x} t\,|f(f(t))|\,dt \text{ for all } x \in [-1, 2].
G
(
x
)
=
∫
−
1
x
t
∣
f
(
f
(
t
))
∣
d
t
for all
x
∈
[
−
1
,
2
]
.
If
lim
x
→
1
F
(
x
)
G
(
x
)
=
1
14
\lim_{x \to 1} \frac{F(x)}{G(x)} = \frac{1}{14}
lim
x
→
1
G
(
x
)
F
(
x
)
=
14
1
, then the value of
f
(
1
2
)
f\left(\frac{1}{2}\right)
f
(
2
1
)
is
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