Let $f:(0,1) \in(0,1)$ be a differentiable function such that $f(x) \neq 0$ for all $x \in(0,1)$ and $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$. Suppose for all $x$,

$\lim\limits_{t \rightarrow x} \frac{\int\limits_0^t \sqrt{1-(f(s))^2} d s-\int\limits_0^x \sqrt{1-(f(s))^2} d s}{f(t)-f(x)}=f(x)$

Then, the value of $f\left(\frac{1}{4}\right)$ belongs to

$\left\{\frac{\sqrt{7}}{4}, \frac{\sqrt{15}}{4}\right\}$

Applying L'Hospital's rule on LHS, we get

$\lim\limits\lim\limits_{t \in x} \frac{\sqrt{1-\{f(t)\}^2}}{f'(t)}=f(x)$

$\Rightarrow \frac{\sqrt{1-(f(x))^2}}{f'(x)}=f(x)$

$\Rightarrow \frac{f(x) f'(x)}{\sqrt{1-(f(x))^2}}=1$

$\Rightarrow \frac{-2 f(x) f'(x)}{\sqrt{1-(f(x))^2}}=-2$

$\Rightarrow \frac{1}{\sqrt{1-(f(x))^2}} d\left\{1-(f(x))^2\right\}=-2$

$\Rightarrow 2 \sqrt{1-(f(x))^2}=-2 x+C$    [On integrating]          .....(i)

Putting $x=\frac{1}{2}$ on both sides and using $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$, we get $C=2$. Putting $C=2$ in (i), we get

$\sqrt{1-(f(x))^2}=-x+1$

$\Rightarrow f(x)=\sqrt{2 x-x^2}$

$\Rightarrow f\left(\frac{1}{4}\right)=\frac{\sqrt{7}}{4}$

Hence, $f\left(\frac{1}{4}\right)$ belongs to $\left\{\frac{\sqrt{7}}{4}, \frac{\sqrt{15}}{4}\right\}$.