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

7

It is given that $f(x)$ is continuous on $R$. Therefore, $F(x)=\int\limits_{-1}^x f(t) d t$ and $G(x)=\int\limits_{-1}^x t|f(f(t))| d t$ are also continuous such that $F'(x)=f(x)$ and $G'(x)=x|f(f(x))|$. Since $f(x)$ is an odd function. Therefore,

$F(1)=\int\limits_{-1}^1 f(t) d t=0 \text { and } G(1)=\int\limits_{-1}^1 t|f(f(t))| d t=0$

Now,

$\lim\limits_{x \rightarrow 1} \frac{F(x)}{G(x)}=\lim\limits_{x \rightarrow 1} \frac{\frac{F(x)-F(1)}{G(x)-G(1)}}{x-1}=\lim\limits_{x \rightarrow 1} \frac{F'(x)}{G'(x)}$

$=\lim\limits_{x \rightarrow 1} \frac{f(x)}{x|f(f(x))|}=\frac{f(1)}{|f(f(1))|}=\frac{1 / 2}{|f(1 / 2)|}=\frac{1}{2|f(1 / 2)|}$

∴  $\lim\limits_{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$

$\Rightarrow \frac{1}{2|f(1 / 2)|}=\frac{1}{14} \Rightarrow|f(1 / 2)|=7 \Rightarrow f(1 / 2)= \pm 7$

Note that $f(1 / 2) \neq-7$ as $f(x)$ vanishes exactly at one point.