Question

Let $f:R\to R$ be a continuous odd function, which vanishes exactly at one point and $f\left(1\right)=\frac{1}{2}.$ Suppose that $F\left(x\right)=\underset{-1}{\overset{x}{\int }}f\left(t\right)dt$ for all $x\in [-1,2]$ and $G\left(x\right)=\underset{-1}{\overset{x}{\int }}t|f\left(f\right(t\left)\right)|dt$ for all $x\in [-1,2]$. If $\underset{x\to 1}{lim}\frac{F\left(x\right)}{G\left(x\right)}=\frac{1}{14}$, then the value of $f\left(\frac{1}{2}\right)$ is

Answer 1

$F\left(1\right)=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=0(\because f(x\left)\text{is an odd function}\right)$
Similarly, $G\left(1\right)=0$

$\underset{x\to 1}{lim}\frac{F\left(x\right)}{G\left(x\right)}=\underset{x\to 1}{lim}\frac{F^{\prime }\left(x\right)}{G^{\prime }\left(x\right)}(\because F(1)=G(1)=0,\text{L-Hospital Rule})$
$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x|f\left(f\right(x\left)\right)|}=\frac{\frac{1}{2}}{|f\left(\frac{1}{2}\right)|}=\frac{1}{14}$
$f\left(\frac{1}{2}\right)=7$