Let f:R→R be a continuous odd function, which vanishes exactly at one point and f(1)=12. Suppose that F(x)=x∫−1f(t)dt for all x∈[−1,2] and G(x)=x∫−1t|f(f(t))|dt for all x∈[−1,2]. If limx→1F(x)G(x)=114, then the value of f(12) is

Let f:R→R be a continuous odd function, which vanishes exactly at on

1.	Let f:R→R be a continuous odd function, which vanishes exactly at one point and f(1)=12. Suppose that F(x)=x∫−1f(t)dt for all x∈[−1,2] and G(x)=x∫−1t\|f(f(t))\|dt for all x∈[−1,2]. If limx→1F(x)G(x)=114, then the value of f(12) is
Answer» Let $f : R \to R$ be a continuous odd function, which vanishes exactly at one point and $f (1) = \frac{1}{2} .$ Suppose that $F (x) = x \int - 1 f (t) d t$ for all $x \in [- 1, 2]$ and $G (x) = x \int - 1 t \| f (f (t)) \| d t$ for all $x \in [- 1, 2]$ . If $lim x \to 1 \frac{F (x)}{G (x)} = \frac{1}{14}$ , then the value of $f (\frac{1}{2})$ is