Let $a,b$ be two real numbers such that $ab < 0$. IF the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a+ib$ lies on the circle $|z-1|=|2z|$, then a possible value of $\frac{1+[a]}{4b}$, where $[t]$ is greatest integer function, is :

$\begin{aligned} & \left|\frac{1+a i}{b+i}\right|=1 \\\\ & |1+i a|=|b+i| \\\\ & a^2+1=b^2+1 \Rightarrow \mathrm{a}=\pm \mathrm{b} \Rightarrow \mathrm{b}=-\mathrm{a} \quad \text { as } \mathrm{ab}<0 \\\\ & (\mathrm{a} +\mathrm{ib}) \text { lies on }|z-1|=|2 z| \\\\ & |a+i b-1|=2|a+i b| \\\\ & (a-1)^2+b^2=4\left(a^2+b^2\right) \\\\ & (a-1)^2=a^2=4\left(2 a^2\right) \\\\ & 1-2 a=6 a^2 \Rightarrow 6 a^2+2 a-1=0 \\\\ & a=\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6} \\\\ & a=\frac{\sqrt{7}-1}{6} \text { and } b=\frac{1-\sqrt{7}}{6}\end{aligned}$ <br/><br/>$$ \begin{aligned} & {[a]=0} \\\\ & \therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right) \end{aligned} $$ <br/><br/>Similarly when $a = {{ - 1 - \sqrt 7 } \over 6}\,\text { and }\,b = {{1 + \sqrt 7 } \over 6}$ <br/><br/>then [$a$] = -1 <br/><br/>$\therefore$ $${{1 + \left[ a \right]} \over {4b}} = {{1 - 1} \over {4 \times {{1 + \sqrt 7 } \over 6}}}$$ = 0